MHT CET · Maths · Three Dimensional Geometry
The equation of the line passing through the point \((-1,3,-2)\) and perpendicular to each of the lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}\) is
- A \(\frac{x+1}{2}=\frac{y-3}{7}=\frac{z+2}{4}\)
- B \(\frac{x+1}{2}=\frac{y+3}{-7}=\frac{z+2}{4}\)
- C \(\frac{x-1}{2}=\frac{y+3}{7}=\frac{z-2}{4}\)
- D \(\frac{x-1}{2}=\frac{y+3}{-7}=\frac{z-2}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{x+1}{2}=\frac{y+3}{-7}=\frac{z+2}{4}\)
Step-by-step Solution
Detailed explanation
The direction ratios of the required line can be obtained by
\(\frac{\mathrm{a}}{2 \times 5-2 \times 3}=\frac{\mathrm{b}}{-3 \times 3-1 \times 5}=\) \(\frac{\mathrm{c}}{1 \times 2-(-3) \times 2} \)
\( \Rightarrow\langle\mathrm{a}, \mathrm{b}, \mathrm{c}\rangle \equiv\langle 4,-14,8\rangle \equiv\langle 2,-7,4\rangle\)
Hence, the equation of required line
\(\begin{aligned} & \frac{x-(-1)}{2}=\frac{y-3}{-7}=\frac{z-(-2)}{4} \\ & \Rightarrow \frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\end{aligned}\)
\(\frac{\mathrm{a}}{2 \times 5-2 \times 3}=\frac{\mathrm{b}}{-3 \times 3-1 \times 5}=\) \(\frac{\mathrm{c}}{1 \times 2-(-3) \times 2} \)
\( \Rightarrow\langle\mathrm{a}, \mathrm{b}, \mathrm{c}\rangle \equiv\langle 4,-14,8\rangle \equiv\langle 2,-7,4\rangle\)
Hence, the equation of required line
\(\begin{aligned} & \frac{x-(-1)}{2}=\frac{y-3}{-7}=\frac{z-(-2)}{4} \\ & \Rightarrow \frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\end{aligned}\)
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