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AP EAMCET · Maths · Three Dimensional Geometry

Equation of the line passing through the intersection of the plane \(x+2 y+3 z=4\) and the line \(\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}\) and parallel to the vector \((2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \times(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})\) is

  1. A \(\frac{x-5}{3}=\frac{y-1}{2}=\frac{z+1}{-7}\)
  2. B \(\frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z-1}{7}\)
  3. C \(\frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z+1}{-7}\)
  4. D \(\frac{x-5}{-3}=\frac{y-1}{2}=\frac{z+1}{7}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z+1}{-7}\)

Step-by-step Solution

Detailed explanation

The general point \(p\) on the line \(\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}=r \quad \text{(Let)} \quad \ldots (i)\) is \(P(2 r+1, r-1,1-r)\). Let the point \(P\) is the intersection of line (i) and the plane \(x+2 y+3 z=4\), so \(2 r+1+2 r-2+3-3 r=4 \Rightarrow r=2\) So,…