Consider the lines L_1:x + 1 \over 3 = y + 2 \over 1 = z + 1 \over 2L_2:x - 2 \over 1 = y + 2 \over 2 = z - 3 \over 3 The unit vector perpendicular to both L_1 and L_2 is :

- \widehat i - 7\widehat j + 5\widehat k \over 5\sqrt 3

- \widehat i + 7\widehat j + 7\widehat k \over \sqrt 99

- \widehat i + 7\widehat j + 5\widehat k \over 5\sqrt 3

7\widehat i - 7\widehat j - \widehat k \over \sqrt 99

Vector Algebra PYQ — JEE Advanced 2008 · Shift 2 · Q27 with Solution

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Correct answer: B

For $L_1:\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ the direction vector is $\vec d_1=3\hat i+\hat j+2\hat k.$

For $L_2:\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$ the direction vector is $\vec d_2=\hat i+2\hat j+3\hat k.$

A vector perpendicular to both direction vectors is given by their cross product:

\begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix}$$ Now expand: $$\vec d_1\times \vec d_2= \hat i(1\cdot 3-2\cdot 2)-\hat j(3\cdot 3-2\cdot 1)+\hat k(3\cdot 2-1\cdot 1)$$ $$=\hat i(3-4)-\hat j(9-2)+\hat k(6-1)$$ $$=-\hat i-7\hat j+5\hat k.$$ So one vector perpendicular to both lines is $$-\hat i-7\hat j+5\hat k.$$ 3. **Convert it into a unit vector** Magnitude: $$\left| -\hat i-7\hat j+5\hat k \right|=\sqrt{(-1)^2+(-7)^2+5^2}$$ $$=\sqrt{1+49+25}=\sqrt{75}=5\sqrt{3}.$$ Hence the unit vector is $$\frac{-\hat i-7\hat j+5\hat k}{5\sqrt3}.$$ 4. **Match with options** This is exactly **Option B**. > Note: its negative would also be perpendicular to both lines, but among the given options, only **B** matches. **Final Answer:** $$\boxed{\frac{-\hat i-7\hat j+5\hat k}{5\sqrt3}}$$