Question

The direction cosines of a normal to the plane passing through (4, 2, 3), (-1, 4, 2) and (3, 2, 1) are

• A. $\frac{-2}{\sqrt{101}}, \frac{3}{\sqrt{101}}, \frac{8}{\sqrt{101}}$
• B. $\frac{-3}{\sqrt{49}}, \frac{2}{\sqrt{49}}, \frac{6}{\sqrt{49}}$
• C. $\frac{-4}{\sqrt{101}}, \frac{-9}{\sqrt{101}}, \frac{2}{\sqrt{101}}$
• D. $\frac{4}{22}, \frac{-12}{22}, \frac{18}{22}$

Hint:

D.C.s are found by dividing direction ratios by the magnitude of the vector.

Answer 1

The Correct Option is C

Step 1: Concept
Normal vector $\vec{n} = \vec{AB} \times \vec{AC}$.

Step 2: Analysis
$\vec{AB} = (-5, 2, -1)$.
$\vec{AC} = (-1, 0, -2)$.

Step 3: Calculation
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 2 & -1 \\ -1 & 0 & -2 \end{vmatrix} = \hat{i}(-4) - \hat{j}(10-1) + \hat{k}(0+2) = (-4, -9, 2)$.
Magnitude $= \sqrt{16 + 81 + 4} = \sqrt{101}$.

Step 4: Conclusion
Direction cosines are $(\frac{-4}{\sqrt{101}}, \frac{-9}{\sqrt{101}}, \frac{2}{\sqrt{101}})$. Final Answer: (C)