Question

If \((1,-2,-2)\) and \((0,2,1)\) are direction ratios of two lines, then the direction cosines of a line perpendicular to both the lines are:

• A. \(\left(\dfrac{1}{3},-\dfrac{1}{3},\dfrac{2}{3}\right)\)
• B. \(\left(\dfrac{2}{3},-\dfrac{1}{3},\dfrac{2}{3}\right)\)
• C. \(\left(-\dfrac{2}{3},-\dfrac{1}{3},\dfrac{2}{3}\right)\)
• D. \(\left(\dfrac{2}{\sqrt{14}},-\dfrac{1}{\sqrt{14}},\dfrac{3}{\sqrt{14}}\right)\)

Hint:

For a line perpendicular to two given lines:

Take the cross product of their direction ratios
Normalize the resulting vector to get direction cosines
Both directions (±) are acceptable

Answer 1

The Correct Option is C

Step 1: Recall the concept. A line perpendicular to two given lines has direction ratios equal to the cross product of their direction ratios. Step 2: Write the given direction ratios. \[ \vec{a} = \langle 1,\,-2,\,-2 \rangle,\quad \vec{b} = \langle 0,\,2,\,1 \rangle \] Step 3: Find the cross product \(\vec{a} \times \vec{b}\). \[ \vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
1 & -2 & -2
0 & 2 & 1 \end{vmatrix} \] \[ = \mathbf{i}((-2)(1)-(-2)(2)) - \mathbf{j}(1\cdot1-(-2)\cdot0) + \mathbf{k}(1\cdot2-(-2)\cdot0) \] \[ = \mathbf{i}(-2+4) - \mathbf{j}(1) + \mathbf{k}(2) \] \[ = \langle 2,\,-1,\,2 \rangle \] Step 4: Convert direction ratios into direction cosines. Magnitude: \[ |\vec{n}| = \sqrt{2^2+(-1)^2+2^2} = \sqrt{9} = 3 \] Direction cosines: \[ \left(\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right) \] The opposite direction is also valid: \[ \left(-\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right) \] Step 5: Match with the given options. Option (C) matches one valid set of direction cosines. Final Answer: \[ \boxed{\left(-\dfrac{2}{3},-\dfrac{1}{3},\dfrac{2}{3}\right)} \]