Question

The vector equation of the plane passing through the intersection of r · (i + j + k) = 1 and r · (i - 2j) = -2, and the point (1, 0, 2) is :

• A. r · (i + 7j + 3k) = 7
• B. r · (i + 7j + 3k) = 7
• C. r · (i - 7j + 3k) = 7/3
• D. r · (i + 7j + 3k) = -7/3

Hint:

Family of planes: $P_1 + \lambda P_2 = 0$. This is the most efficient way to find a plane through the intersection line of two other planes.

Answer 1

The Correct Option is A

Step 1: Equation of family of planes: $(x+y+z-1) + \lambda(x-2y+2) = 0$.
Step 2: Passes through $(1, 0, 2)$: $(1+0+2-1) + \lambda(1-0+2) = 0 \Rightarrow 2 + 3\lambda = 0 \Rightarrow \lambda = -2/3$.
Step 3: Equation: $(x+y+z-1) - \frac{2}{3}(x-2y+2) = 0$.
Step 4: $3x+3y+3z-3 - 2x + 4y - 4 = 0 \Rightarrow x + 7y + 3z = 7$.
Step 5: Vector form: $\vec{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$.