A non-zero vector \(\vec{a}\) is parallel to the line of intersection of the plane determined by the vectors \(\hat{i}\), \(\hat{i} + \hat{j}\) and the plane determined by the vectors \(\hat{i} - \hat{j}\), \(\hat{i}+ \hat{j}\). The angle between \(\vec{a}\) and the vector \(\hat{i}-2\hat{j}+\hat{k}\) is :
Answer & explanation
Correct answer: option 1
Since \(\vec{a}\) is parallel to the line of intersection of the two planes, it is parallel to both the planes. Thus, \(\vec{a}\) is perpendicular to the normals of both the planes.
Hence, \(\vec{a}\).[\(\hat{i}\)x(\(\hat{i}\) + \(\hat{j}\))] = 0 and \(\vec{a}\).[(\(\hat{i} - \hat{j}\))x(\(\hat{j}+\hat{k} )\)] = 0
\(\vec{a}.\hat{k} = 0\) and \(\vec{}.[(\hat{i}+\hat{j})] = 0\) so that \(\vec{a}\) is in the direction of \(\hat{i}-\hat{j}\)
Hence, \(\vec{a} = \pm \frac{1}{\sqrt{2}} (\hat{i}-\hat{j})\). If \(\theta\) is the angle between the given vector and \(\vec{a}\),
\(\cos {\theta} = \pm \frac{1}{\sqrt{2}} [\hat{i}-\hat{j}].\frac{1}{3} [\hat{i} - 2\hat{j} + 2\hat{k}] = \pm \frac{1}{\sqrt{2}}\)
⇒ \(\theta = \frac{\pi}{4}\) or \(\frac{3\pi}{4}\)