TS EAMCET · Maths · Three Dimensional Geometry
The position vector of a point \(P\) is \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) and \(\mathbf{a}=-\hat{\mathbf{i}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) are two vectors which determine a plane \(\pi\). The equation of a line through \(P\) normal to \(\mathbf{b}\) and lying on the plane \(\pi\) is
- A \(r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(-\hat{i}+5 \hat{j}-2 \hat{k})\)
- B \(r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}+\hat{j}+\hat{k})\)
- C \(r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(-2 \hat{i}-\hat{j}+3 \hat{k})\)
- D \(r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(-3 \hat{i}+4 \hat{j}-5 \hat{k})\)
Answer & Solution
Correct Answer
(A) \(r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(-\hat{i}+5 \hat{j}-2 \hat{k})\)
Step-by-step Solution
Detailed explanation
Given, Position vector of a point \(P\) is \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})\) Equation of plane, \((\mathbf{r}-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(\mathbf{a} \times \mathbf{b})=0\) \(\therefore\) Normal of plane…
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