3D Geometry
Equation of Plane
Grade 12
Question:
<p>The cartesian equations of the plane which passes through the point <i>(5, 2, −4)</i> and perpendicular to the line with direction ratios <i>2, 3, −1</i> is</p>
<p>(a) <i>2x + 2y − z = 20</i></p>
<p>(b) <i>2x + y + z = 10</i></p>
<p>(c) <i>2x − 3y + z = 10</i></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the vector form of a plane passing through a point with a given normal vector, then convert to cartesian form.
Solution: We have, the position vector of point (5, 2, −4) say $\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}$ and the normal vector $\vec{n}$ perpendicular to the plane as $\vec{n} = 2\hat{i} + 3\hat{j} - \hat{k}$ Therefore, the vector equation of the plane is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$ $[\vec{r} - (5\hat{i} + 2\hat{j} - 4\hat{k})] \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$ Transforming into cartesian form: $[(x - 5)\hat{i} + (y - 2)\hat{j} + (z + 4)\hat{k}] \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$ $2(x - 5) + 3(y - 2) - 1(z + 4) = 0$ $2x + 3y - z = 20$ ∴ Answer is (d) None of these, since the correct answer is $2x + 3y - z = 20$.
Correct Answer: D