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Three-dimensional Geometry involves the mathematics of shapes in 3D space and involves 3 coordinates which are x-coordinate, y-coordinate and z-coordinate. During a 3D space, three parameters are required to seek out the precise location to some extent.

Three-dimensional geometry plays a serious role with roughly 4 to 5 questions in **JEE Main Question Paper** covering a weightage of 16- 20%. Looking at the previous year trends, a total of 14- 16 questions are from Class 12th syllabus out of which 3-4 questions are asked from three dimensional geometry. **Check ****JEE Main Mathematics Syllabus**

3D Geometry is sort of complex and conceptual. It also includes complex numbers alongside some parts of geometry. You’ll easily score well from this section if you understand certain complex concepts of 3D properly. Attempt to develop a strong hold on basics and focus more on understanding this unit through visualization.

Some of the important subtopics from Three Dimensional Geometry in **JEE Main** exam include Coordinates of a point in space, the distance between two points; Section formula, direction ratios and direction cosines, the angle between two intersecting lines; Skew lines, the shortest distance between them and its equation; Equations of a line and a plane in different forms, the intersection of a line and a plane, coplanar lines.

### JEE Main Test Series Three Dimensional Geometry: Previous Year Questions and Solutions

**Question 1: **The projection of any line on coordinate axes be, respectively 3, 4, 5 then its length is ______.

**Solution:**

Let the line segment be AB, then as given AB cos α = 3, AB cos β = 4, AB cos γ = 5

⇒ AB2 (cos2α + cos2β + cos2γ) = 32 + 42 + 52

AB = √[9 + 16 + 25] = 5√2,

Where α, β and γ are the angles made by the line with the axes?

**Question 2: **The angle between two diagonals of a cube will be _____.

**Solution:**

Let the cube be of side ‘a’ and O (0, 0, 0), D (a, a, a), B (0, a, 0), G (a, 0, a).

Then the equations of OD and BG are x/a = y/a = z/a and x/a = [y − a]/[−a] = z/a, respectively.

Hence, the angle between OD and BG is

cos^{−1} \frac{a^2 − a^2 + a^2}{\sqrt{3}a^2 \times \sqrt{3}a^2} = cos^{−1} (\frac{1 }{ 3})cos−13a2×3a2a2−a2+a2=cos−1(31)

**Question 3:** The coordinates of the foot of perpendicular drawn from point P (1, 0, 3) to the line joining the points A (4, 7, 1) and B (3, 5, 3) is _______.

**Solution:**

Let D be the foot of perpendicular drawn from P (1, 0, 3) on the line AB joining (4, 7, 1) and (3, 5, 3).

If D divides AB in ratio λ : 1, then

D = [(\frac{3 \lambda + 4}{\lambda + 1}), (\frac{5 \lambda + 7}{\lambda + 1}), (\frac{3 \lambda + 1}{\lambda + 1})]D=[(λ+13λ+4),(λ+15λ+7),(λ+13λ+1)] …..(i)

Direction ratios of PD are 2λ + 3, 5λ + 7, −2.

Direction ratios of AB are −1, −2, 2 [Because PD⊥AB]

− (2λ + 3) −2 (5λ + 7) − 4 = 0

λ = −7 / 4

Putting the value of λ in (i), we get the point D (5/3, 7/3, 17/3).

**Question 4:** A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The coordinates of each of the points of intersection are given by __________.

**Solution:**

Let the two lines be AB and CD having equations

x / 1 = [y + a] / 1 = z / 1 = λ and [x + a] / [2] = y / 1 = z / 1 = μ then

P = (λ, λ − a, λ) and Q = (2μ − a, μ, μ)

So, according to question, [λ − 2μ + a] / [2] = [λ − a − μ] / [1] = [λ − μ] / [2]

μ = a and λ = 3a

Therefore, P = (3a, 2a, 3a) and [(x − 2)2 +(y − 3)2 + (z − 4)2].

**Question 5:** The distance of the point (1, 2, 3) from the plane x − y + z = 5 measured parallel to the line x / 2 = y / 3 = z / −6, is __________.

**Solution:**

Direction cosines of line = (2 / 7, 3 / 7, −6 / 7)

Now, x′ = 1 + [2r /7], y′ = −2+ [3r / 7] and z′ = 3 − [6r / 7]

Therefore, (1 + [2r / 7]) − (−2 + [3r / 7]) + (3 − [6r / 7])

= 5

⇒ r = 1

**Question 6:** A point moves so that the sum of its distances from the points (4, 0, 0) and (4, 0, 0) remains 10. The locus of the point is _______.

**Solution:**

\sqrt{(x − 4)^2 + y^2 + z^2 } + \sqrt{(x + 4)^2 + y^2 + z^2} = 10(x−4)2 +y2 +z2 +(x+4)2 +y2 +z2=10

2 (x2 + y2 + z2) + 2 \sqrt{[(x − 4)^2 + y^2 + z^2] [(x + 4)^2 + y^2 + z^2]}[(x−4)2 +y2+z2][(x+4)2+y2+z2] = 100 − 32 = 68

⇒ (x2 + y2 + z2 − 34)2 = [(x − 4)2 + y2 + z2] [(x + 4)2 + y2 + z2]

⇒ (x2 + y2 + z2)2 − 68 (x2 + y2 + z2) + (34)2 = [(x2 + y2 + z2 + 16) − 8x] [(x2 + y2 + z2 + 16) + 8x]

= (x2 + y2 + z2 + 16)2 − 64x2

= (x2 + y2 + z2) + 32 (x2 + y2 + z2) − 64x2 + (16)2

⇒ 9x2 + 25y2 + 25z2 − 225 = 0

**Question 7: **The centre of the sphere passes through four points (0, 0, 0), (0, 2, 0), (1, 0, 0) and (0, 0, 4) is ________.

**Solution:**

Let the equation of sphere be x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0

Because it passes through origin (0, 0, 0), d = 0 and also, it passes through (0, 2, 0)

4 + 4v = 0 ⇒ v = −1

Also, it passes through (1, 0, 0)

1 + 2u = 0 ⇒ u = − 1/2

And it passes through (0, 0, 4)

16 + 8w ⇒ w = −2

Centre (−u, −v, −w)=(1 / 2, 1, 2)

**Question 8: **Co-ordinate of a point equidistant from the points (0,0,0), (a, 0, 0), (0, b, 0), (0, 0, c) is ___________.

**Solution:**

The required point is the centre of the sphere through the given points.

Let the equation of sphere be x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 …..(i)

Sphere (i) is passing through (0, 0, 0), (a, 0, 0), (0, b, 0) and (0, 0, c), hence, d = 0

a2 + 2ua = 0 ⇒ u = −a/2

b2 + 2vb = 0 ⇒ v = −b/2

c2 + 2wc = 0 ⇒ w = −c/2

Therefore, the centre of sphere is (a/2, b/2, c/2), which is also the required point.

**Question 9:** If the plane 2ax − 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres x2 + y2 + z2 + 6x − 8y − 2z = 13 and x2 + y2 + z2 − 10x + 4y − 2z = 8, then a equals ____________.

**Solution:**

S1 = x2 + y2 + z2 + 6x − 8y − 2z = 13, C1 = (−3, 4, 1)

S2 = x2 + y2 + z2 − 10x + 4y − 2z = 8, C2 = (5, −2, 1)

So, mid point of C1 C2 (say P) = P(\frac{5 − 3}{2}, \frac{4 − 2}{2}, \frac{1+1}{2})25−3,24−2,21+1) = P(1, 1, 1)

Now the plane 2ax − 3ay + 4az + 6 = 0 passes through the point P,

So, 2a (1) − 3a (1) + 4a (1) + 6 = 0 = 2a −3a + 4a + 6 = 0

3a + 6 = 0

3a = −6

⇒a = −2

**Question 10:** The radius of the sphere x + 2y + 2z = 15 and x2 + y2 + z2 − 2y − 4z = 11 is ___________.

**Solution:**

Equation of sphere is, x2 + y2 + z2 − 2y − 4z = 11

Centre of sphere = (0, 1, 2) and radius of sphere = 4

Let the centre of the circle be (α, β, γ).

The direction ratios of a line joining from centre of the sphere to the centre of the circle are

(α − 0, β − 1, γ − 2) or (α, β − 1, γ − 2)

But, this line is normal at plane x + 2y + 2z = 15

α / 1 = [β − 1]/2 = [γ − 2]/2 = k

α = k, β = 2k + 1, γ = 2k + 2 [Because, centre of circle lies on x + 2y + 2z = 15]

k + 2 (2k + 1) + 2 (2k + 2) = 15

⇒ k = 1

So, the centre of circle = (1, 3, 4)

Therefore, Radius of circle = √[(Radius of sphere)2 − (Length of joining line of centre)2] \sqrt{4^2 − (1 − 0)^2 + (3 − 1)^2 + (4 − 2)^2 } = \sqrt{16-9}42−(1−0)2+(3−1)2+(4−2)2=16−9

= √7

**Question 11:** The point at which the line joining the points (2, 3, 1) and (3, 4, 5) intersects the plane 2x + y + z = 7 is _________.

**Solution:**

Ratio = \frac{2 (2) + (−3) (1) + (1) (1) − 7} {2 (3) + (−4) (1) + (−5) (1) − 7} = \frac{−(−5)}{− 10}2(3)+(−4)(1)+(−5)(1)−72(2)+(−3)(1)+(1)(1)−7=−10−(−5) = −(1 / 2)

Hence, x = \frac{2 (2) − 3 (1)}{1}12(2)−3(1) = 1, y = \frac{−3 (2) − (−4)}{1}1−3(2)−(−4) = −2 and z = \frac{1 (2) − (−5)}{1}11(2)−(−5) = 7.

Therefore, P (1, −2, 7).

Trick : As (1, 2, 7) and (1, 2, 7) satisfy the equation 2x + y + z = 7, but the point (1, 2, 7) is collinear with (2, 3, 1) and (3, 4, 5).

Note: If a point divides the joining of two points in some particular ratio, then this point must be collinear with the given points.

**Question 12:** The point where the line \frac{x − 1}{2} = \frac{y − 2}{−3} = \frac{z + 3}{4}2x−1=−3y−2=4z+3 meets the plane 2x + 4y − z = 1, is ___________.

**Solution:**

Let point be (a, b, c), then 2a + 4b − c = 1 …..(i) and a = 2k + 1,b = −3k + 2 and c = 4k − 3,

(where k is constant)

Substituting these values in (i), we get

2 (2k + 1) + 4 (−3k + 2) − (4k − 3) = 1

⇒ k = 1

Hence, the required point is (3, 1, 1).

Trick: The point must satisfy the lines and plane.

**Question 13: **The equation of the straight line passing through the point (a, b, c) and parallel to the z-axis, is ________.

**Solution:**

The line through (a,b,c) is \frac{x − a}{l} = \frac{y − b}{m} = \frac{z -c}{n}lx−a=my−b=nz−c ….(i)

Since the line is parallel to z-axis, therefore any point on this line will be of the form (a, b, z1). Also, any point on line (i) is (lr + a, mr + b, nr + c).

Hence, lr + a = a and mr + b = b ⇒ l = m = 0

Hence, the line will be \frac{x − a}{0} = \frac{y − b}{0} = \frac{z -c}{1}0x−a=0y−b=1z−c

**Question 14: **The acute angle between the line joining the points (2, 1, 3), (3, 1, 7) and a line parallel to \frac{x − 1}{3} = \frac{y }{4} = \frac{z + 3}{5}3x−1=4y=5z+3 through the point (1, 0, 4) is ____________.

**Solution:**

Direction ratio of the line joining the point (2, 1, −3), (−3, 1, 7) are (a1, b1, c1) ⇒ (−3 −2, 1 − 1, 7 − (−3))

⇒ (−5, 0, 10)

Direction ratio of the line parallel to line \frac{x − 1}{3} = \frac{y }{4} = \frac{z + 3}{5}3x−1=4y=5z+3 are (a2, b2, c2)

⇒ (3, 4, 5)

Angle between two lines,

cos θ = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{(a_1^2 + b_1^2 + c_1^2 )} \times \sqrt{(a_2^2 + b_2^2 + c_2^2 )}}(a12+b12+c12 )×(a22+b22+c22 )a1a2+b1b2+c1c2

cos θ = \frac{-5 \times 3 + 0 \times 4 + 10 \times 5}{\sqrt{(25 + 0 + 100)} \times \sqrt{(9 + 16 + 25 )}}(25+0+100)×(9+16+25 )−5×3+0×4+10×5

cos θ = \frac{35}{25 \sqrt{10}}251035

⇒ θ = cos−1 (\frac{7}{5 \sqrt{10}}5107)

**Question 15:** If l1, m1, n1 and l2, m2, n2 are the direction cosines of two perpendicular lines, then the direction cosine of the line which is perpendicular to both the lines, will be ______.

**Solution:**

Let lines are l1x + m1y + n1z + d = 0 ..(i) and

l2x + m2y + n2z + d = 0 …..(ii)

If lx +my + nz + d = 0 is perpendicular to (i) and (ii), then, ll1 + mm1 + nn1 = 0, ll2 + mm2 + nn2 = 0

⇒ \frac{l }{m_1n_2− m_2n_1} = \frac{m}{n_1l_2 − l_1n_2} = \frac{n }{l_1m_2 − l_2m_1} = dm1n2−m2n1l=n1l2−l1n2m=l1m2−l2m1n=d

Therefore, direction cosines are (m1n2 − m2n1), (n1l2 − l1n2), (l1m2 − l2m1).

### JEE Main Three Dimensional Geometry: Quick Formulas

Here is a list of brief formulas used in the Three Dimensional Geometry.

**1. Vector representation of a point:** Position vector of a point P(x, y, z) is x\hat{i}+y\hat{j}+z\hat{k}xi^+yj^+zk^

**2.Distanceformula:** \sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}(x1−x2)2+(y1−y2)2+(z1−z2)2 , AB = \left | \vec{OB} -\vec{OA}\right |∣∣∣∣OB−OA∣∣∣∣

**3. Distance of P from coordinate axes:** PA = \sqrt{(y^{2}+z^{2})}(y2+z2)

PB = \sqrt{(z^{2}+x^{2})}(z2+x2)

PC = \sqrt{(x^{2}+y^{2})}(x2+y2)

**4. Section Formula:**

x = \frac{mx_{2}+nx_{1}}{m+n}m+nmx2+nx1

y = \frac{my_{2}+ny_{1}}{m+n}m+nmy2+ny1

z = \frac{mz_{2}+nz_{1}}{m+n}m+nmz2+nz1

Midpoint:

x = \frac{x_{1}+x_{2}}{2}2x1+x2

y = \frac{y_{1}+y_{2}}{2}2y1+y2

z = \frac{z_{1}+z_{2}}{2}2z1+z2

**5. Centroid of a triangle:**

G = \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3},\frac{z_{1}+z_{2}+z_{3}}{3} \right )(3x1+x2+x3,3y1+y2+y3,3z1+z2+z3)

**6. Incentre of triangle ABC:** \left ( \frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\frac{ay_{1}+by_{2}+cy_{3}}{a+b+c},\frac{az_{1}+bz_{2}+cz_{3}}{a+b+c} \right )(a+b+cax1+bx2+cx3,a+b+cay1+by2+cy3,a+b+caz1+bz2+cz3)

**7. Centroid of a tetrahedron:** \left ( \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\frac{y_{1}+y_{2}+y_{3}+y_{4}}{4},\frac{z_{1}+z_{2}+z_{3}+z_{4}}{4} \right )(4x1+x2+x3+x4,4y1+y2+y3+y4,4z1+z2+z3+z4)

**8. Direction cosines and direction ratios:**

(i) Direction cosines: let α, β, γ be the angles which a directed line makes with the positive directions of the axes of x, y and z respectively, then cos α, cos β and cos γ are called the direction cosines of the line. The direction cosines are usually denoted by (l, m, n).

Therefore l = cos α, m = cos β, n = cos γ.

(ii) l2+m2+n2 = 1

(iii) If a, b, c are the direction ratios of any line L then a\hat{i}+b\hat{j}+c\hat{k}ai^+bj^+ck^ will be a vector parallel to the line L.

(iv) If l, m, n are the direction cosines of any line L then l\hat{i}+m\hat{j}+n\hat{k}li^+mj^+nk^ is a unit vector parallel to the line L.

(v) If l, m, n be the direction cosines and a, b, c be the direction ratios of a vector, then

l = \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}a2+b2+c2a

m = \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}a2+b2+c2b

n = \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}a2+b2+c2c

or l = \frac{-a}{\sqrt{a^{2}+b^{2}+c^{2}}}a2+b2+c2−a,

m = \frac{-b}{\sqrt{a^{2}+b^{2}+c^{2}}}a2+b2+c2−b

n = \frac{-c}{\sqrt{a^{2}+b^{2}+c^{2}}}a2+b2+c2−c

(vi) If OP = r, the direction cosines of OP are l, m, n then the coordinates of P are ( lr, mr, nr).

If the direction cosines of the line AB are l, m, n, |AB| = r and the coordinates of A is (x1, y1, z1) then the coordinates of B are given as (x1+rl, y1+rm, z1+rn).

(vii) If the coordinates P and Q are (x1, y1, z1) and (x2, y2, z2) then the direction ratios of line PQ are a = x2-x1 , b = y2-y1 and c = z2-z1 and the direction cosines of line PQ are l = \frac{x_{2}-x_{1}}{\left | PQ \right |}∣PQ∣x2−x1, m = \frac{y_{2}-y_{1}}{\left | PQ \right |}∣PQ∣y2−y1 and n = \frac{z_{2}-z_{1}}{\left | PQ \right |}∣PQ∣z2−z1

(vii) Direction cosines of the x-axis is (1, 0, 0).

Direction cosines of the y-axis is (0, 1, 0).

Direction cosines of the z-axis is (0, 0, 1).

**9. Angle between two line segments:**

cos θ = \left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \right |∣∣∣∣∣a12+b12+c12a22+b22+c22a1a2+b1b2+c1c2∣∣∣∣∣

The line will be perpendicular if a1a2+b1b2+c1c2 = 0, and parallel if a1/a2 = b1/b2 = c1/c2.

**10. Projection of a line segment on a line:**

If P(x1, y1, z1) and Q(x2, y2, z2) then the projection of PQ on a line having direction cosines l, m, n is |l(x2-x1)+m(y2-y1)+n(z2-z1)|

**11. Equation of a plane: General form:** ax+by+cz+d = 0, where a, b, c are not all zero, a, b, c, d ∈ R.

(i) Normal form: lx+my+nz = p

(ii) Plane through the point (x1, y1, z1): a(x-x1)+b(y-y1)+c(z-z1) = 0

(iii) Intercept form: \frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1ax+by+cz=1

(iv) vector form: (\vec{r}-\vec{a} ).\vec{n}= 0(r−a).n=0 or \vec{r}.\: \vec{n}= \vec{a}.\: \vec{n}r.n=a.n

(v) Planes parallel to the axes :

(a) plane parallel to X-axis is by+cz+d = 0

(b) plane parallel to Y-axis is ax+cz+d = 0

(c) plane parallel to Z-axis is ax+by+d = 0

(vi) Plane through origin: Equation of the plane passing through the origin is ax+by+cz = 0.

(vii) Transformation of the equation of a plane to the normal form: ax+by+cz-d = 0in normal form is \frac{ax}{\pm \sqrt{a^{2}+b^{2}+c^{2}}}+\frac{by}{\pm \sqrt{a^{2}+b^{2}+c^{2}}}+\frac{cz}{\pm \sqrt{a^{2}+b^{2}+c^{2}}} = \frac{d}{\pm \sqrt{a^{2}+b^{2}+c^{2}}}±a2+b2+c2ax+±a2+b2+c2by+±a2+b2+c2cz=±a2+b2+c2d

(viii) Any plane parallel to the given plane ax+by+cz+d = 0 is ax+by+cz+λ = 0.

Distance between ax+by+cz+d1 = 0 and ax+by+cz+d2 = 0 is \frac{\left | d_{1}-d_{2} \right |}{\sqrt{a^{2}+b^{2}+c^{2}}}a2+b2+c2∣d1−d2∣

(ix) A plane ax+by+cz+d = 0 divides the line segment joining (x1, y1, z1) and (x2, y2, z2) in the ratio \left ( -\frac{ax_{1}+by_{1}+cz_{1}+d}{ax_{2}+by_{2}+cz_{2}+d}\right )(−ax2+by2+cz2+dax1+by1+cz1+d)

(x) Coplanarity of four points: The points A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) are coplanar if \begin{vmatrix} x_{2}-x_{1}& y_{2}-y_{1} & z_{2}-z_{1}\\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \\ x_{4}-x_{1}& y_{4}-y_{1} & z_{4}-z_{1} \end{vmatrix}= 0∣∣∣∣∣∣∣x2−x1x3−x1x4−x1y2−y1y3−y1y4−y1z2−z1z3−z1z4−z1∣∣∣∣∣∣∣=0

**12. A point and a plane:**

(i) distance of the point (x’, y’, z’) from the plane ax+by+cz+d = 0 is given by \frac{ax^{‘}+by^{‘}+cz^{‘}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}a2+b2+c2ax‘+by‘+cz‘+d

(ii) Length of the perpendicular from a point \vec{a}a to the plane \vec{r}.\vec{n} = dr.n=d is given by p = \frac{\left |\vec{a} .\: \vec{n} -d\right |}{\left | \vec{n} \right |}∣n∣∣a.n−d∣.

(iii) Foot (x’, y’, z’) of perpendicular drawn from the point (x1, y1, z1) to the plane ax+by+cz+d = 0 is given by \frac{x^{‘}-x_{1}}{a} = \frac{y^{‘}-y_{1}}{b} =\frac{z^{‘}-z_{1}}{c}= -\frac{ax_{1}+by_{1}+cz_{1}+d}{a^{2}+b^{2}+c^{2}}ax‘−x1=by‘−y1=cz‘−z1=−a2+b2+c2ax1+by1+cz1+d

(iv) To find image of a point with respect to a plane:

Let P (x1, y1, z1) be a given point and ax+by+cz+d = 0 is given plane. Let (x’, y’, z’) is the image point. Then \frac{x^{‘}-x_{1}}{a} = \frac{y^{‘}-y_{1}}{b} =\frac{z^{‘}-z_{1}}{c}= -2\frac{(ax_{1}+by_{1}+cz_{1}+d)}{a^{2}+b^{2}+c^{2}}ax‘−x1=by‘−y1=cz‘−z1=−2a2+b2+c2(ax1+by1+cz1+d).

(v) The distance between two parallel planes ax+by+cz+d = 0 and ax+by+cz+d’ = 0 is given by \frac{\left | d-d{}’ \right |}{\sqrt{a^{2}+b^{2}+c^{2}}}a2+b2+c2∣d−d’∣

**13. Angle between two planes:**

(i) \cos \theta =\left | \frac{aa{}’+bb{}’+cc{}’}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{a{}’^{2}+b{}’^{2}+c{}’^{2}}}\right |cosθ=∣∣∣∣a2+b2+c2a’2+b’2+c’2aa’+bb’+cc’∣∣∣∣

(ii) Planes are perpendicular if aa’+bb’+cc’ = 0 and parallel if a/a’ = b/b’ = c/c’.

(iii) The angle θ between the planes \vec{r}.\vec{n_{1}} = d_{1}r.n1=d1 and \vec{r}.\vec{n_{2}} = d_{2}r.n2=d2 is given by cos θ = \frac{\vec{n_{1}}.\vec{n_{2}} }{\left | \vec{n_{1}} \right |.\left | \vec{n_{2}} \right |}∣n1∣.∣n2∣n1.n2

(iv) Planes are perpendicular if \vec{n_{1}}.\vec{n_{2}}n1.n2 = 0 and planes are parallel if \vec{n_{1}}=\lambda \vec{n_{2}}n1=λn2, λ is a scalar.

**14. Angle bisectors:**

(i) The equations of a planes bisecting the angle between two given planes a1x+b1y+c1z+d1 = 0 and a2x+b2y+c2z+d2 = 0 are \frac{a_{1}x+b_{1}y+c_{1}z+d_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}a12+b12+c12a1x+b1y+c1z+d1 = \pm \frac{a_{2}x+b_{2}y+c_{2}z+d_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}±a22+b22+c22a2x+b2y+c2z+d2

(ii) Bisector of acute or obtuse angle: First make both the constant terms positive. Then a1a2+b1b2+c1c2 > 0 ⇒ origin lies on obtuse angle.

a1a2+b1b2+c1c2 < 0 ⇒ origin lies on acute angle.

**15. Area of a triangle:**

Let A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) be the vertices of a triangle, then Δ = \sqrt{\Delta_{x}^{2}+\Delta _{y}^{2}+\Delta _{z}^{2}}Δx2+Δy2+Δz2 where \Delta _{x} = \frac{1}{2}\begin{vmatrix} y_{1} & z_{1} & 1\\ y_{2} &z_{2} & 1\\ y_{3}& z_{3} & 1 \end{vmatrix}Δx=21∣∣∣∣∣∣∣y1y2y3z1z2z3111∣∣∣∣∣∣∣, \Delta _{y} = \frac{1}{2}\begin{vmatrix} z_{1} & x_{1} & 1\\ z_{2} &x_{2} & 1\\ z_{3}& x_{3} & 1 \end{vmatrix}Δy=21∣∣∣∣∣∣∣z1z2z3x1x2x3111∣∣∣∣∣∣∣ and \Delta _{z} = \frac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2} &y_{2} & 1\\ x_{3}& y_{3} & 1 \end{vmatrix}Δz=21∣∣∣∣∣∣∣x1x2x3y1y2y3111∣∣∣∣∣∣∣

Vector method: from two vector \vec{AB}AB and \vec{AC}AC the area is given by \frac{1}{2}\left |\vec{AB} \times \vec{AC} \right |21∣∣∣∣AB×AC∣∣∣∣

**16. Volume of a tetrahedron:**

Volume of a tetrahedron with vertices A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) is given by V = \frac{1}{6}\begin{vmatrix} x_{1} & y_{1} & z_{1} & 1\\ x_{2} & y_{2} & z_{2} & 1\\ x_{3}& y_{3} & z_{3} & 1\\ x_{4}& y_{4} & z_{4} & 1 \end{vmatrix}61∣∣∣∣∣∣∣∣∣x1x2x3x4y1y2y3y4z1z2z3z41111∣∣∣∣∣∣∣∣∣

**17. Equation of a line:**

(i) A straight line is the intersection of two planes. It is represented by two planes a1x+b1y+c1z+d1 = 0 and a2x+b2y+c2z+d2 = 0 .

(ii) Symmetric form: \frac{(x-x_{1})}{a} = \frac{(y-y_{1})}{b}= \frac{(z-z_{1})}{c} = ra(x−x1)=b(y−y1)=c(z−z1)=r

(iii) vector equation:\vec{r} = \vec{a}+\lambda \vec{b}r=a+λb

(iv) Reduction of cartesian form of equation of a line to vector form and vice versa \frac{x-x_{1}}{a} = \frac{y-y_{1}}{b}= \frac{z-z_{1}}{c}ax−x1=by−y1=cz−z1 ⇔\vec{r} = (x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k})+\lambda (a\hat{i}+b\hat{j}+c\hat{k})r=(x1i^+y1j^+z1k^)+λ(ai^+bj^+ck^)

**18. To find image of a point with respect to a line:**

Let L = \frac{x-x_{2}}{a} = \frac{y-y_{2}}{b}= \frac{z-z_{2}}{c}ax−x2=by−y2=cz−z2 is a given line.

Let (x’, y’, z’) is the image of the point (x1, y1, z1) with respect to the line L. Then

(i) a(x1-x’)+b(y1-y’)+C(z1-z’) = 0

(ii) \frac{\frac{x_{1}+x{}’}{2}-x_{2}}{a}= \frac{\frac{y_{1}+y{}’}{2}-y_{2}}{b}= \frac{\frac{z_{1}+z{}’}{2}-z_{2}}{c} = \lambdaa2x1+x’−x2=b2y1+y’−y2=c2z1+z’−z2=λ

From (ii) get the value of x’, y’, z’ in terms of λ as x’= 2aλ+2x2-x1, y’= 2bλ-2y2-y1,

z’= 2cλ+2z2-z1 . Then put the values of x’, y’, z’ in 9i) get λ and substitute the value of λ to get (x’, y’,z’).

**19. Angle between a line and a plane:**

(i) If θ is the angle between a line \frac{x-x_{1}}{l}= \frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}lx−x1=my−y1=nz−z1 and the plane ax+by+cz+d = 0, then \sin \theta = \left | \frac{al+bm+cn}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{l^{2}+m^{2}+n^{2}}} \right |sinθ=∣∣∣∣a2+b2+c2l2+m2+n2al+bm+cn∣∣∣∣

(ii) Vector form : If θ is the angle between a line \vec{r}= \vec{a}+\lambda \vec{b}r=a+λb and \vec{r}.\vec{n }= dr.n=d then sin θ = \frac{\vec{b}.\vec{n}}{\left | \vec{b} \right |\left | \vec{n} \right |}∣b∣∣n∣b.n

(iii) condition for perpendicularity l/a = m/b = n/c , \vec{b}\times \vec{n}= 0b×n=0

(iv) condition for parallel : al+bm+cn = o, \vec{b}.\vec{n}= 0b.n=0

**20. Condition for a line to lie in a plane:**

(i) cartesian form: Line \frac{x-x_{1}}{l}= \frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}lx−x1=my−y1=nz−z1 would lie in a plane ax+by+cz+d = 0, if ax1+by1+cz1+d = 0 and al+bm+cn = 0

(ii) Vector form: Line \vec{r}= \vec{a}+\lambda br=a+λb would lie in the plane \vec{r}.\vec{n}= dr.n=d if \vec{b}.\vec{n}= 0b.n=0 and \vec{a}.\vec{n}= da.n=d

**21. Skew lines:**

(i) The straight lines which are not parallel and non-coplanar are called skew lines.

If \Delta = \begin{vmatrix} \alpha ^{‘} -\alpha & \beta {}’-\beta &\gamma {}’-\gamma \\ l & m &n \\ l’& m’&n’ \end{vmatrix} \neq 0Δ=∣∣∣∣∣∣∣α‘−αll’β’−βmm’γ’−γnn’∣∣∣∣∣∣∣=0 , then the lines are skew.

(ii) Shortest distance:

SD = \frac{\begin{vmatrix} \alpha ^{‘} -\alpha & \beta {}’-\beta &\gamma {}’-\gamma \\ l & m &n \\ l’& m’&n’ \end{vmatrix} }{\sqrt{\sum (mn’-m’n)^{2}}}∑(mn’−m’n)2∣∣∣∣∣∣α‘−αll’β’−βmm’γ’−γnn’∣∣∣∣∣∣

(iii) Vector form: For lines \vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}}r=a1+λb1 and \vec{r} = \vec{a_{2}}+\lambda \vec{b_{2}}r=a2+λb2 to be skew (\vec{b_{1}}\times \vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})\neq 0(b1×b2).(a2−a1)=0

(iv) Shortest distance between lines \vec{r} = \vec{a}+\lambda \vec{b}r=a+λb

and \vec{r} = \vec{a_{2}}+\mu \vec{b}r=a2+μb is d= \left | \frac{(\vec{a_{2}}-\vec{a}_{1})\times \vec{b}}{\left | \vec{b} \right |} \right |d=∣∣∣∣∣∣b∣(a2−a1)×b∣∣∣∣∣

**22. Family of planes:**

(i) Any plane through the intersection of a1x+b1y+c1z+d1 = 0 and a2x+b2y+c2z+d2 = 0 is a1x+b1y+c1z+d1+λ(a2x+b2y+c2z+d2) = 0

(ii) The equation of plane passing through the intersection of the planes \vec{r}.\vec{n_{1}}= d_{1}r.n1=d1 and \vec{r}.\vec{n_{2}}= d_{2}r.n2=d2 is \vec{r}.(\vec{n_{1}}+\lambda \vec{n_{2}})=d_{1}+ \lambda d_{2}r.(n1+λn2)=d1+λd2 where λ is an arbitrary scalar.

**23. Sphere:**

General equation of a sphere is x2+y2+z2+2ux+2vy+2wz+d = 0. (-u, -v, -w) is the centre and \sqrt{u^{2}+v^{2}+w^{2}-d}u2+v2+w2−d is the radius of the sphere.