Question

If \[\vec A\] = \[\hat i\] + \[\hat j\] + 3\[\hat k\], \[\vec B\] = -\[\hat i\] + \[\hat j\] + 4\[\hat k\] and \[\vec C\] = 2\[\hat i\] - 2\[\hat j\] - 8\[\hat k\], then the angle between the vectors \[\hat P\] = \[\vec A\] + \[\vec B\] + \[\vec C\] and \[\vec Q\] = (\[\vec A\] × \[\vec B\]) is (in degree) ______.

Options

• 0°

• 45°

• 90°

• 60°

Answer 1

If \[\vec A\] = \[\hat i\] + \[\hat j\] + 3\[\hat k\], \[\vec B\] = -\[\hat i\] + \[\hat j\] + 4\[\hat k\] and \[\vec C\] = 2\[\hat i\] - 2\[\hat j\] - 8\[\hat k\], then the angle between the vectors \[\hat P\] = \[\vec A\] + \[\vec B\] + \[\vec C\] and \[\vec Q\] = (\[\vec A\] × \[\vec B\]) is (in degree) 90°.

Explanation:

\[\vec P\] = \[\vec A\] + \[\vec B\] + \[\vec C\] = \[\hat{\mathrm{i}}+\hat{\mathrm{j}}+3\hat{\mathrm{k}}-\hat{\mathrm{i}}+\hat{\mathrm{j}}+4\hat{\mathrm{k}}+2\hat{\mathrm{i}}-2\hat{\mathrm{j}}-8\hat{\mathrm{k}}\]

\[\vec{\mathrm{P}}=2\hat{\mathrm{i}}-\hat{\mathrm{k}}\]

\[\vec Q\] = \[\vec A\] × \[\vec B\] = \[\begin{vmatrix} \hat{1} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & 3 \\ -1 & 1 & 4 \end{vmatrix}\] = \[\hat{\mathrm{i}}(4-3)-\hat{\mathrm{j}}(4+3)+\hat{\mathrm{k}}(1+1)\]

\[\vec{\mathrm{Q}}=\hat{\mathrm{i}}-7\hat{\mathrm{j}}+2\hat{\mathrm{k}}\]

Angle between \[\vec P\] and \[\vec Q\] is given by cos θ = \[\frac{\vec{\mathrm{P}}\cdot\vec{\mathrm{Q}}}{|\vec{\mathrm{P}}||\vec{\mathrm{Q}}|}\]

= \[\frac{(2\hat{\mathrm{i}}-\hat{\mathrm{k}})\cdot(\hat{\mathrm{i}}-7\hat{\mathrm{j}}+2\hat{\mathrm{k}})}{\sqrt{\left(2\right)^{2}+\left(0\right)^{2}+\left(1\right)^{2}}\cdot\sqrt{\left(1\right)^{2}+\left(7\right)^{2}+\left(2\right)^{2}}}\] = 0

\[\therefore\] θ = 90°