Question

Which of the following is are correct regarding vectors Ā i+j and B = 2i – 3j

* Multiple options may be correct

•Angle between A and B is acute

•Angle between A +B and A is acute

•k is a unit vector perpendicular to A

•(i-j)/✓2 is a anit vector perpendicular to A​

Answer 1

Answer:

1 and 2 are correct......

Answer 2

Answer: The correct options are b, c &d.

The angle between A and A+B is acute.

k^ is a unit vector perpendicular to A.

(i^- j^)/✓2 is a unit vector perpendicular to A​.

Explanation:

A=  i^+j^

B = 2i^ – 3j^

A.B = (i^+j^) . (2i^ – 3j^)

Let θ be the angle between A and B.

a) |A||B| cosθ = (i^+j^) . (2i^ – 3j^)

    |A||B| cosθ = 2 - 3 = -1    

=> cosθ = $\frac{-1}{\sqrt[]{13} \sqrt{2} }$ = $\frac{-1}{\sqrt[]{26} }$ = $\frac{-1}{5.1}$ = -0.196

=> θ = $cos^{-1}$ (-0.196) = 99.7296°

=> θ > 90°

Hence, the angle between A and B is obtuse.

So, option 1 is false.

b)  Now, A + B = i^+j^ +  2i^ – 3j^  = 3i^ - 2j^

     A= i^+j^

Similarly, the angle between A+B and A is α.

α =   $cos^{-1}$( $\frac{(A + B).A}{|A| |A+B|}$)

=> α =  $cos^{-1}$( $\frac{1}{\sqrt{2} \sqrt{36} }$)

=> α =  $cos^{-1}$( 0.176) =  79.8631

=> α < 90°

Hence, the angle between A and A+B is acute.

c)  Now, k^. A = k^ . (i^ + j^)

          => k^. A = 0

This implies that k^ is a unit vector perpendicular to A.

d)   (i^-j^)/✓2. A =  (i^-j^)/✓2 . ( i^+j^)

                         = $\frac{1}{\sqrt{2} }$ -  $\frac{1}{\sqrt{2} }$

                         = 0

This implies that (i^- j^)/✓2 is a unit vector perpendicular to A​.