MHT CET · Maths · Three Dimensional Geometry
The unit vectors perpendicular to the plane determined by the points \(\mathrm{A}(1,-1,2) \mathrm{~B}(2,0,-1) \mathrm{~C}(0,2,1)\) is
- A \(\pm\left(\frac{3 \hat{i}+\hat{j}+\hat{k}}{\sqrt{11}}\right)\)
- B \(\pm\left(\frac{-\hat{i}+2 \hat{j}+\hat{k}}{\sqrt{6}}\right)\)
- C \(\pm\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)\)
- D \(\pm\left(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right)\)
Answer & Solution
Correct Answer
(C) \(\pm\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)\)
Step-by-step Solution
Detailed explanation
\(\vec{AB} = (2-1)\hat{i} + (0-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k}\) \(\vec{AC} = (0-1)\hat{i} + (2-(-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}\)
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