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
- A \(\frac{x-5}{3}=\frac{y-1}{2}=\frac{z+1}{-7}\)
- B \(\frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z-1}{7}\)
- C \(\frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z+1}{-7}\)
- D \(\frac{x-5}{-3}=\frac{y-1}{2}=\frac{z+1}{7}\)
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,…
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