MHT CET · Maths · Three Dimensional Geometry
The unit vector perpendicular to the plane \(4 x-3 y+12 z=15\) is
- A \(\frac{4 \hat{\imath}+3 \hat{\jmath}+12 \hat{k}}{13}\)
- B \(\frac{4 \hat{\imath}-3 \hat{\jmath}+12 \hat{k}}{13}\)
- C \(\frac{-4 \hat{\imath}+3 \hat{\jmath}+12 \hat{k}}{13}\)
- D \(\frac{-4 \hat{\imath}-3 \hat{\jmath}+12 \hat{k}}{13}\)
Answer & Solution
Correct Answer
(B) \(\frac{4 \hat{\imath}-3 \hat{\jmath}+12 \hat{k}}{13}\)
Step-by-step Solution
Detailed explanation
The unit vector perpendicular to plane \(4 x-3 y+12 z=15\) is \(\frac{4 \hat{i}-3 \hat{j}+12 \hat{k}}{\sqrt{16+9+144}}=\frac{4 \hat{i}-3 \hat{j}+12 \hat{k}}{13}\)
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