Question

If \( \vec{p} = 4\vec{i} - \vec{j} + \vec{k} \) is a point and \( \vec{q} = 9\vec{i} - 2\vec{j} + 6\vec{k} \) is a vector, then the perpendicular distance of origin from the plane passing through \( \vec{p} \) and perpendicular to \( \vec{q} \) is:

• A. \( 4 \)
• B. \( 3\sqrt{2} \)
• C. \( 9 \)
• D. \( 11 \)

Hint:

Use the plane equation \( \vec{n} \cdot (\vec{r} - \vec{r_0}) = 0 \) and apply distance formula from a point to a plane.

Answer 1

The Correct Option is A

The equation of the plane passing through point \( \vec{p} \) and normal to vector \( \vec{q} \) is: \[ \vec{q} \cdot (\vec{r} - \vec{p}) = 0 \] Substitute values: \[ (9, -2, 6) \cdot ((x, y, z) - (4, -1, 1)) = 0 \Rightarrow 9(x - 4) - 2(y + 1) + 6(z - 1) = 0 \Rightarrow 9x - 2y + 6z = 36 + 2 + 6 = 44 \] Now, distance from origin to the plane: \[ \frac{|9(0) - 2(0) + 6(0) - 44|}{\sqrt{9^2 + (-2)^2 + 6^2}} = \frac{44}{\sqrt{81 + 4 + 36}} = \frac{44}{\sqrt{121}} = 4 \]