Question

Find the vector of magnitude 5 units and in the opposite direction to 2i + 3j - 6k.

Answer 1

Step-by-step explanation:

let

a=2i+3j-6k

magnitude =5

Required vector =

magnitude × unit vector in opposite direction

unit vector in opposite direction=

= -(2i+3j-6k)/√2²+3³+6²

=-(2i+3j-6K)/√49

=-(2i+3j-6k)/7

required vector= -5(2i+3j-6k)/7

Answer 2

The vector of magnitude 5 units is given by

$5(-\frac{2}{7}i-\frac{3}{7}j+\frac{6}{7}k)$

Step-by-step explanation:

Let vector

$\vec{a}=2i+3j-6k$

Magnitude of vector r=xi+yj+zk is given by

$\mid r\mid=\sqrt{x^2+y^2+z^2}$

Using the formula

$\mid a\mid=\sqrt{2^2+3^2+(-6)^2}=7$

Unit vector=$\hat{a}=\frac{\vec{a}}{\mid a\mid}$

Using the formula

$\hat{a}=\frac{2i+3j-6k}{7}$

We are given that the direction of vector is opposite to given vector therefore, the vector

$\hat{a}=-(\frac{2i+3j-6k}{7})=\frac{-2}{7}i-\frac{3}{7}j+\frac{6}{7}k$

Magnitude=5

Therefore, the vector of magnitude 5 units is given by

$5\hat{a}=5(-\frac{2}{7}i-\frac{3}{7}j+\frac{6}{7}k)$

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