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Vector indicate two particular points that are directed from one place to another. The direction of a vector is defined as the direction in which it acts. The angle made by the vector with the x-axis is known as the direction of a vector. The counterclockwise rotation of the angle of the vector around its tail due east determines the vector's direction. Vector can also be defined as the difference between velocity and speed. Velocity has both magnitude and direction, hence, it is a vector quantity, but speed only has magnitude, so it is a scalar quantity.
Read also: Unit vector, Vector product of two vectors
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Key Terms: Direction, vector, magnitude, force, quadrant, direction of a vector, unit vector
Also Read: Addition of vectors
Direction of Vector
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An angle of rotation of the vector about its tail from north, south, east, or west is expressed as the direction of the vector. The direction of a vector is the direction along which a vector acts. The direction of the vector is expressed through the direction of the arrowhead. The direction of the vector also explains the intensity at which the object is moving beside the direction towards which the object is moving.
Read More: Determinant Formula
The direction of vector is denoted by \(\to\) a = |a|^ . |a| = |a|a^. |a| denotes the magnitude of the vector, ^a denoted the direction of the vector.
The direction of a vector formula is related to the slope of a line. The slope of a line that passes through the origin and a point (x, y) is y/x. If θ is the angle made by this line, then its slope is tan θ, i.e., tan θ = y/x.
Hence, θ = tan-1 (y/x). Thus, the direction of a vector (x, y) is found using the formula tan-1 (y/x) but while calculating this angle, the quadrant in which (x, y) lies also should be considered.
Read More: Applications of the Integrals
Steps to find the direction of a vector (x, y):
Find α using α = tan-1 |y/x|.
The quadrant on which (x,y) lies are important to find the direction of the vector \(\ominus\).
| The quadrant on which (x,y) lies | \(\ominus\) (in degrees) |
|---|---|
| 1 | α |
| 2 | 180o - α |
| 3 | 180o + α |
| 4 | 360o - α |
The video below explains this:
Types of Vectors Detailed Video Explanation:
Also Read: Angle between two vectors
Types of vectors
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The different types of vectors are:
- zero vector: a vector whose starting and ending points meet at the same point is known as zero vector.
- unit vector: a vector whose magnitude is 1 unit is known as a unit vector.
- coinitial vectors: two or more vectors having a common starting point is known as a coinitial vector.
- collinear vectors: when two or more vectors are parallel to the same line irrespective of the direction and magnitude. it is known as collinear vectors.
- equal vectors: if two vectors have the same direction and magnitude without a common starting point. it is known as equal vectors.
- negative of a vector: if the magnitude of a vector is similar to that of the given vector but has an opposite direction. then it is known as the negative of a vector.
Read also: Resultant vector formula, Displacement vector
Things to Remember
- Direction and magnitude are expressed as a vector.
- A vector is expressed by a line with an arrow on the top and a fixed point at the other end.
- Velocity is a kind of vector
- Force vector is the direction in which force is applied.
- The direction of the vector formula and the slope of a line are interrelated.
- The direction of a vector (x,y) = tan-1 (y/x)
- Vector (x,y) = (x2 – x1 , y2 – y1)
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Sample Questions
Ques. If P is the initial point which is marked at (5,2) and Q is the endpoint which is marked at (4,3), find the direction of the vector PQ. (3 marks)
Ans: we know, (x1.y1) = (5,2)
(x2.y2) = (4,3)
\(\ominus\) = tan-1{(y2- y1)/(x2 – x1)
\(\ominus\) = tan-1 \(\frac{3-2}{4-5}\)
\(\ominus\)= tan-1 (-1)
\(\ominus\) = -45\(\circ\) or 135\(\circ\)
Ques. Using the vector formula, find the direction of vector (1 , -√3). (2 marks)
Ans: (x,y) = (1, -√3)
α = tan-1 | y/x |
α = tan-1 |-\(\sqrt{}3\)/1| = tan-1 √3 = 60\(\circ\)
we know that (1, -√3) lies in quadrant 4
\(\ominus\) = 360 – α = 360 – 60 = 300\(\circ\)
Ques. The starting point of the vector is at (1,3) and ends at (-4,-2). Find the direction of the vector. (3 marks)
Ans: given, (x1,y1) = (1,3)
(x2, y2) = (-4,-2)
(x,y) = (x2 – x1 , y2 – y1 )
= (-4-1, -2-3) = (-5,-5)
Gα = tan-1 |-5/-5|
= tan-11 = 45\(\circ\)
(-5,-5) lies in quadrant 3
The direction of the given vector is,
\(\ominus\) = 180 + α = 180 +45
=225\(\circ\)
Ques. P is the initial point of the vector with points (1,4) and Q with the points (3,9). Find the direction of the vector. (1 marks)
Ans: the coordinates of the vector PQ
(x,y) = (3-1,9-4) =(2,5)
\(\ominus\) = tan-1 |5/2|
= 68.2\(\circ\)
Ques. The stating point of the vector PQ is P (2,3) and the end point Q is at (5,8). Find the direction. (1 marks)
Ans: tan \(\ominus\) = (y2 – y1)/(x2 – x1)
= \(\frac{8-3}{5-2}\) = \(\frac{5}{3}\)
\(\ominus\) = tan-1 (\(\frac{5}{3}\))
= 59\(\circ\)
Ques. Find out the direction of a vector which is directed from origin to the (3,5) (1 marks)
Ans: a = x= 3
Bb = y = 5
\(\ominus\) = tan – 1 (a/b)
\(\ominus\)= tan– 1 (3/5)
\(\ominus\) = 30.9\(\circ\)
Ques. Given u = (3,-2) and v = (-1,4) , find two new vectors \(\to\)u- v and \(\to\) u + v. (1 marks)
Ans: u + v = (3,-2) +(-1,4)
= (3 + (-1), -2 + 4)
= (2,2)
u + (-v) = (3,-2) + (1,-4)
= (3+1,-2 +(-4))
=(4,-6)
Ques. Vector \(\to\) a have a magnitude of 1 and vector \(\to\) b have a magnitude of 2. Find the angle between the two angles if \(\to\)a. \(\to\)b = 1. (1 marks)
Ans: as we already know , \(\to\)a. \(\to\)b = 1
| \(\to\)a| = 1
|\(\to\)b| = 2
Thus, \(\ominus\) = cos-1 (\(\frac{{\to}a.{\to}b}{ |{\to}a||{\to}b|}\))
= cos-1 =( \(\frac{1}{2}\))
= \(\frac{\pi}{3}\)
Ques. If vector \(\to\)a and vector \(\to\)b is given as |\(\to\)a| =2 , |\(\to\)b|= 3 and \(\to\)a. \(\to\)b = 4. Find |\(\to\)a. \(\to\)b|. (2 marks)
Ques. Calculate |\(\to\)x| when \(\to\)a = unit vector, (\(\to\)x - \(\to\)a).(\(\to\)x + \(\to\)a ) = 8. (1 marks)
Ans: we know, \(\to\)a = unit vector and |\(\to\)a|= 1
(\(\to\)x - \(\to\)a). (\(\to\)x + \(\to\)a) = 8
Or, |\(\to\)x|2 – 1 = 8
i.e, |\(\to\)x|2 = 9
= |\(\to\)x|= 3
Ques. Calculate the unit vector in the same direction as v = (-5, 12). (1 marks)
Ans: magnitude |v| = \(\sqrt{((-5)^2 + (12)^2)}\)
= √(25+144)
= \(\sqrt{169}\)
= 13
=1
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