TS EAMCET · Maths · Vector Algebra
A bisector of the angle between the normals of the planes \(4 x+3 y=5\) and \(x+2 y+2 z=4\) is along the vector
- A \((17 \hat{i}+9 \hat{j}-12 \hat{k})\)
- B (b) \((17 \hat{i}-9 \hat{j}+12 \hat{k})\)
- C \((17 \hat{i}-\hat{j}+10 \hat{k})\)
- D \((7 \hat{i}-\hat{j}-10 \hat{k})\)
Answer & Solution
Correct Answer
(D) \((7 \hat{i}-\hat{j}-10 \hat{k})\)
Step-by-step Solution
Detailed explanation
Given planes are \(4 x+3 y-5=0 \& x+2 y+2 z-4=0\). Equation of plane which bisects the angle between the given planes. \(\begin{aligned} & \frac{4 x+3 y-5}{\sqrt{16+9}}= \pm \frac{x+2 y+2 z-4}{\sqrt{1+4+4}} \\ & \frac{4 x+3 y-5}{5}= \pm \frac{x+2 y+2 z-4}{3} \end{aligned}\) Take…
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