MHT CET · Maths · Three Dimensional Geometry
The Cartesian equation of the line passing through the point \((-3,0,1)\) and perpendicular to vectors \(\hat{i}-2 \widehat{j}+\widehat{k}\) and \(2 \hat{i}+\widehat{j}-\widehat{k}\) is
- A \(\frac{x+3}{1}=\frac{y}{3}=\frac{z-1}{-5}\)
- B \(\frac{x+3}{-1}=\frac{y}{3}=\frac{z-1}{5}\)
- C \(\frac{x+3}{1}=\frac{y}{3}=\frac{z-1}{5}\)
- D \(\frac{x+3}{1}=\frac{y}{-3}=\frac{z-1}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{x+3}{1}=\frac{y}{3}=\frac{z-1}{5}\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{x}+3}{(-2) \times(-1)-1 \times 1}=\frac{\mathrm{y}-0}{2 \times 1-1 \times(-1)}=\) \(\frac{\mathrm{z}-1}{1 \times 1-2 \times(-2)}\)
\(\Rightarrow \frac{\mathrm{x}+3}{1}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}-1}{5}\)
\(\Rightarrow \frac{\mathrm{x}+3}{1}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}-1}{5}\)
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