Original JEE Main and CUET-style questions on vectors and three-dimensional geometry, with a worked solution and the usual trap for each. Tap an option, or type a numerical answer, to check it.
Learn the chapters first: Class 11, Introduction to Three Dimensional Geometry · Class 12, Vector Algebra · Class 12, Three Dimensional Geometry.
The angle between \(\vec a = \hat i - \hat j\) and \(\vec b = \hat i - \hat k\) is
\(\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \dfrac{1}{\sqrt2\sqrt2} = \dfrac12\), so \(\theta = \dfrac{\pi}{3}\).
The area of the parallelogram with adjacent sides \(\vec a = 3\hat i + \hat j - \hat k\) and \(\vec b = \hat i - \hat j + 2\hat k\) is
\(\vec a \times \vec b = (1 \cdot 2 - (-1)(-1))\hat i - (3 \cdot 2 - (-1) \cdot 1)\hat j + (3(-1) - 1 \cdot 1)\hat k = \hat i - 7\hat j - 4\hat k\).
Area \(= |\vec a \times \vec b| = \sqrt{1 + 49 + 16} = \sqrt{66}\).
Trap: \(\tfrac12|\vec a \times \vec b|\) is the area of the triangle, not the parallelogram.
Find the shortest distance between the lines \(\vec r = (\hat i + 2\hat j + 3\hat k) + t\,\hat i\) and \(\vec r = s\,\hat j\).
Directions \(\hat i\) and \(\hat j\); \(\hat i \times \hat j = \hat k\). The vector between the given points is \(-(\hat i + 2\hat j + 3\hat k)\).
Distance \(= \dfrac{|(-\hat i - 2\hat j - 3\hat k)\cdot\hat k|}{|\hat k|} = 3\).
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