AP EAMCET · Maths · Three Dimensional Geometry
If the line joining the points \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\) meets the plane that passes through the point \(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\) and parallel to the vectors \(3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) and \(3 \hat{\mathbf{i}}-\hat{\mathbf{k}}\) at, \(P\), then the position vector of the point \(P\) is
- A \(-27 \hat{\mathbf{i}}+\hat{\mathbf{j}}+14 \hat{\mathbf{k}}\)
- B \(29 \hat{\mathbf{i}}+\hat{\mathbf{j}}-14 \hat{\mathbf{k}}\)
- C \(-14 \hat{\mathbf{i}}+89 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
- D \(2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-7 \hat{\mathbf{k}}\)
Answer & Solution
Correct Answer
(B) \(29 \hat{\mathbf{i}}+\hat{\mathbf{j}}-14 \hat{\mathbf{k}}\)
Step-by-step Solution
Detailed explanation
Equation of line joining points \((\hat{\mathbf{i}}+\hat{\mathbf{j}})\) and \((3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})\) in cartesian form is \[ \frac{x-1}{2}=\frac{y-1}{0}=\frac{z}{-1}=r(\text { let }) \] Then, point on line Eq. (i) is \(p(2 r+1,1,-r)\). Now,…
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