TS EAMCET · Maths · Three Dimensional Geometry
If \(\mathbf{e}\) is a unit vector perpendicular to the plane determined by the points \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\). If \(\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\), then the projection vector of \(\mathbf{a}\) on \(\mathbf{e}\) is
- A \(\frac{11}{14}(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})\)
- B \(\frac{1}{3}(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\)
- C \(\frac{1}{7}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})\)
- D \(\frac{1}{\sqrt{14}}(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}})\)
Answer & Solution
Correct Answer
(A) \(\frac{11}{14}(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})\)
Step-by-step Solution
Detailed explanation
Plane passing through \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}), \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\). So, point in cartesian form are \((2,1,1),(1,-1,1)\) and \((-1,1,-1)\). Equation of…
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