Question

The unit vector parallel to the resultant of the vectors vector a is equal to 4 icap + 3 j cap + 6 k cap and b vector is equal to minus icap + 3 j cap - k cap is

Answer 1

Answer: The correct answer is $\frac{(3i+6j-5k)}{\sqrt{70}}$.

Explanation:

Find the resultant of the vectors a and b by addition.

R= a+b

Put a= 4i+3j+6k and b= - i + 3 j - k.

R= 4i+3j+6k+(- i + 3 j - k)

R= 3i+6j-5k

Find the magnitude of the resultant vector R.

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

Put x= 3i, y= 6j and z= -5k.

$R=\sqrt{3^{2}+6^{2}+(-5)^{2} }$

$R=\sqrt{9+36+25}$

$R=\sqrt{70}$

Calculate the unit vector.

$unit vector R= \frac{R}{magnitude\ of R}$

Put R= 3i+6j-5k and $R=\sqrt{70}$.

$unit vector R= \frac{(3i+6j-5k)}{\sqrt{70}}$

Therefore, the unit vector of the resultant of the vectors a and b is $\frac{(3i+6j-5k)}{\sqrt{70}}$.

Answer 2

Given:

a=4i+3j+6k

b=-i+3j-k

To find:

Unit vector parallel to the resultant of the vector a and b.

Solution:

a=4i+3j+6k

b=-i+3j-k

c=a+b=4i+3j+6k+(-i+3j-k)=3i+6j+5k

$|c|=|a+b|=\sqrt{x^2+y^2+z^2}$

$|c|=\sqrt{3^2+6^2+5^2}$

$|c|=\sqrt{70}$

Unit vector,$\hat{c}=\frac{c}{|c|}$

Using the formula

$\hat{c}=\frac{3i+6j+5k}{\sqrt{70}}$

Hence, the unit vector parallel to the resultant of the vector a and b is given by

$\frac{3i+6j+5k}{\sqrt{70}}$