Question

Find the angle between p=2i+j+3k AMD Q=3i-2j+k

Answer 1

Given:

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

To Find:

$\sf Angle \ between \ \overrightarrow{P} \ and \ \overrightarrow{Q} \ (\theta)$

Answer:

$\rm \overrightarrow{P}. \overrightarrow{Q} = (2\hat{i} +\hat{j} + 3 \hat{k}). (3\hat{i} - 2 \hat{j} + \hat{k}) \\ \\ \rm \overrightarrow{P}. \overrightarrow{Q} = (2 \times 3) + (1 \times ( - 2) + (3 \times 1) \\ \\ \rm \overrightarrow{P}. \overrightarrow{Q} = 6 - 2 + 3 \\ \\ \rm \overrightarrow{P}. \overrightarrow{Q} = 7$

$\rm | \overrightarrow{P} | = \sqrt{ {2}^{2} + {1}^{2} + {3}^{2} } \\ \\ \rm | \overrightarrow{P} | = \sqrt{4 + 1 + 9} \\ \\ \rm | \overrightarrow{P} | = \sqrt{14} \\ \\ \\ \rm | \overrightarrow{Q} | = \sqrt{ { 3}^{2} + {(-2)}^{2} + {1}^{2} } \\ \\ \rm | \overrightarrow{Q} | = \sqrt{9+ 4+ 1} \\ \\ \rm | \overrightarrow{Q} | = \sqrt{14}$

Let $\rm \theta$ be the angle between $\rm \overrightarrow{P}$ & $\rm \overrightarrow{Q}$

$\rm \implies \overrightarrow{P}. \overrightarrow{Q} = | \overrightarrow{P} | | \overrightarrow{Q} | cos \theta \\ \\ \rm \implies cos \theta = \dfrac{\overrightarrow{P}. \overrightarrow{Q} }{| \overrightarrow{P} | | \overrightarrow{Q} |} \\ \\ \rm \implies \theta = {cos}^{ - 1} ( \dfrac{\overrightarrow{P}. \overrightarrow{Q} }{| \overrightarrow{P} | | \overrightarrow{Q} |} ) \\ \\ \rm \implies \theta = {cos}^{ - 1} ( \dfrac{7 }{ \sqrt{14} \times \sqrt{14} } ) \\ \\ \rm \implies \theta = {cos}^{ - 1} ( \dfrac{7 }{ 14} ) \\ \\ \rm \implies \theta = {cos}^{ - 1} ( \dfrac{1 }{ 2} ) \\ \\ \rm \implies \theta = {60}^{ \circ}$

$\therefore$ $\boxed{\mathfrak{Angle \ between \ \overrightarrow{P} \ and \ \overrightarrow{Q} \ (\theta) = 60\degree}}$

Answer 2

$\huge{\boxed{\sf QuEsTIoN}}$

☯ Find the angle between p=2i+j+3k AMD Q=3i-2j+k

$\huge{\boxed{\sf Answer}}$

P = 2i + j + 3k ; Q = 3i - 2j + k

Now angle between ā and b be

➭ cos-¹ ( 6 - 2 + 3 )

(√14 × √14)

➭ cos-¹ (7/14)

➭ cos-¹ (1/2)

➭ cos 60°

☯ Hence Verified