Question

The angle between the vectors P = 3i+j+2k and Q=i-2i+3k is ​

Answer 1

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Answer 2

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

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




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

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

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

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

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

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

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




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

$\sf{\overrightarrow{P} . \overrightarrow{Q} = (3)(1) + (1)(- 2) + (2)(3)}$

= 3 - 2 + 6

= 7




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

As we know that,

$\sf{\overrightarrow{P} . \overrightarrow{Q} = PQ cos \theta}$

$\succ$ 7 = (√14)(√14) cos θ

$\succ$ 7 = 14 cos θ

$\succ$ cos θ = 1 / 2

$\succ$ cos θ = cos 60°

On comparing both sides, we get

θ = 60°




Hence, $<u>the angle between two given vectors is</u>$ $<I><u><b> 60°.</I></b></u>$