MHT CET · Maths · Vector Algebra
A plane is parallel to two lines direction ratios are \((1,0,-1)\) and \((-1,1,0)\) and it contains the point \((1,1,1)\). If it cuts the co-ordinate axes at A, B, C, then the volume of the tetrahedron \(\mathrm{OABC}\) is cu. units.
- A \(9\)
- B \(27\)
- C \(\frac{9}{4}\)
- D \(\frac{9}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{9}{2}\)
Step-by-step Solution
Detailed explanation
Direction ration of normal to the plane can be obtained by
\(\frac{\mathrm{a}}{0 \times 0-1 \times(-1)}=\frac{\mathrm{b}}{(-1) \times(-1)-1 \times 0}=\frac{\mathrm{c}}{1 \times 1-(-1) \times 0} \)
\( \Rightarrow\langle\mathrm{a}, \mathrm{b}, \mathrm{c}\rangle \equiv\langle 1,1,1\rangle\)
Hence equation of the plane \(1(x-1)+1(y-1)+1(z-1)=0\)
\(\begin{aligned} & \Rightarrow x+y+z=3 \\ & \Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1\end{aligned}\)
\(\Rightarrow\) The plane intersects the co-ordinate axes at \((3,0,0),(0,3,0)\) and \((0,0,3)\)
Hence, volume of tetrahedron OABC is
\(\frac{\mathrm{a}}{0 \times 0-1 \times(-1)}=\frac{\mathrm{b}}{(-1) \times(-1)-1 \times 0}=\frac{\mathrm{c}}{1 \times 1-(-1) \times 0} \)
\( \Rightarrow\langle\mathrm{a}, \mathrm{b}, \mathrm{c}\rangle \equiv\langle 1,1,1\rangle\)
Hence equation of the plane \(1(x-1)+1(y-1)+1(z-1)=0\)
\(\begin{aligned} & \Rightarrow x+y+z=3 \\ & \Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1\end{aligned}\)
\(\Rightarrow\) The plane intersects the co-ordinate axes at \((3,0,0),(0,3,0)\) and \((0,0,3)\)
Hence, volume of tetrahedron OABC is
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