Question

The vectors A= 6i+9j-3k and

B=2i+3j-k

a) Parallel b) anti-parallel c)

Perpendicular d) identical​

Answer 1

$\sf{ \overrightarrow{A} = 6 \hat{i} + 9 \hat{j} - 3 \hat{k}}$

$\sf{\overrightarrow{B} = 2 \hat{i} + 3 \hat{j} - \hat{k}}$




$<b>Step I :</b>$ $<u>Find the magnitude of both the vectors.</u>$

|$\sf{\overrightarrow{A}}$| $\sf{= {\sqrt{(6)^2 + (9)^2 + (- 3)^2}} }$

$\sf{A = {\sqrt{36 + 81 + 9}}}$

$\sf{A = {\sqrt{126}} = 3 {\sqrt{14}}}$

|$\sf{\overrightarrow{B}}$| $\sf{= {\sqrt{(2)^2 + (3)^2 + (- 1)^2}} }$

$\sf{B = {\sqrt{4 + 9 + 1}}}$

$\sf{B = {\sqrt{14}}}$




$<b>Step II :</b>$ $<u>Determine the dot product of both vectors.</u>$

$\sf{\overrightarrow{A} . \overrightarrow{B} = (6)(2) + (9)(3) + (- 3)(- 1)}$

= 12 + 27 + 3

= 42




$<b>Step III :</b>$ $<u>Determine the angle</u>$

As we know that,

$\sf{\overrightarrow{A} . \overrightarrow{B} = AB cos \theta}$

$\succ$ 42 = (3√14)(√14) cos θ

$\succ$ 42 = 3 * 14 cos θ

$\succ$ cos θ = 1

$\succ$ cos θ = cos 0°

On comparing both sides, we get

θ = 0°

This shows that both the vectors are parallel.




Hence, $<u>the correct option is</u>$ $<I><u><b> a) Parallel.</I></b></u>$

Answer 2

Please see the attachment