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COMEDK · Maths · 34. Three Dimensional Geometry

A vector perpendicular to the plane containing the points \(A(1,-1,2), B(2,0,-1), C(0,2,1)\) is

  1. A \(8 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)
  2. B \(4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}\)
  3. C \(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\)
  4. D \(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(8 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)

Step-by-step Solution

Detailed explanation

We have, \(A(1,-1,2), B(2,0,-1)\) and \(C(0,2,1)\) \(\therefore \quad \mathbf{A B}=\{2-1) \mathbf{i}+(0-(-1)) \mathbf{j}+(-1-2) k\) \(=\mathbf{i}+\mathbf{j}-3 \mathbf{k}\) and \(A C=(0-1) \mathbf{i}+(2-(-1)) \mathbf{j}+(1-2) k\) \(=-\mathbf{i}+3 \mathbf{j}-\mathrm{k}\) Now,…