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AP EAMCET · Maths · Three Dimensional Geometry

The cartesian equation of a line \(2 x-3=3 y+1=5-6 z\). The vector equation of the line passing through the point \((7,-5,0)\) and parallel to the given line is

  1. A \(r=(5 \hat{i}-7 \hat{j})+\lambda(3 \hat{i}+2 \hat{j}-\hat{k})\)
  2. B \(r=(7 \hat{i}+5 \hat{j})+\lambda(3 \hat{i}-2 \hat{j}+\hat{k})\)
  3. C \(r=(7 \hat{i}-5 \hat{j})+\lambda(3 \hat{i}+2 \hat{j}-\hat{k})\)
  4. D \(r=(-5 \hat{i}+7 \hat{j})+\lambda(-3 \hat{i}-2 \hat{j}-\hat{k})\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(r=(7 \hat{i}-5 \hat{j})+\lambda(3 \hat{i}+2 \hat{j}-\hat{k})\)

Step-by-step Solution

Detailed explanation

Equation of given line is \[ \begin{aligned} & 2 x-3=3 y+1=5-6 z \\ & \frac{x-\frac{3}{2}}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{5}{6}}{-1} \end{aligned} \] \(\therefore\) Equation of line passes through point \((7,-5,0)\) and having paralleI vector…