Question

A plane P is parallel to two lines whose direction rations are –2, 1, –3 and –1, 2, –2 and it contains the point (2, 2, –2). Let P intersect the co-ordinate axes at the points A, B, C making the intercepts α, β, γ. If V is the volume of the tetrahedron OABC, where O is the origin and p = α + β + γ, then the ordered pair (V, p) is equal to :

• A. (48,-13)
• B. (24,-13)
• C. (48,11)
• D. (24,-5)

Answer 1

The Correct Option is B

Assume \(\begin{array}{l} \overrightarrow{a}_1=\left(-2,1,-3\right) \end{array}\) and\( \begin{array}{l} \overrightarrow{a}_2=\left(-1,2,-2\right)\end{array}\)Vector normal to plane

So, \(\begin{array}{l} \overline{n}=\overrightarrow{a}_1\times \overrightarrow{a}_2\end{array}\)⇒\(\begin{array}{l} \overline{n}=\left(4,-1,-3\right) \end{array}\)

\(\begin{array}{l}\text{Plane through}\ (2, 2, -2)\ \text{and normal to}\ \overline{n}\end{array}\)

\(\begin{array}{l}\left(x – 2, y – 2, z + 2\right) \cdot \left(4, -1, -3\right) = 0\\ \Rightarrow 4x – y – 3z = 12\end{array}\)

\(\begin{array}{l} \Rightarrow\ \frac{x}{3}+\frac{y}{-12}+\frac{z}{-4}=1\end{array}\)

\(Intercepts \space α, β, γ\space are \space 3, –12, –4 \)

\(P = α + β + γ = – 13\)

\(\begin{array}{l} V=\frac{1}{6}\times 3\times 12\times 4=24 \end{array}\)