MHT CET · Maths · Three Dimensional Geometry

The length of perpendicular drawn from the point \(2 \hat{i}-\hat{j}+5 \hat{k}\) to the line \(\overline{\mathrm{r}}=(11 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})+\lambda(10 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-11 \hat{\mathrm{k}})\) is

A \(\sqrt{14}\) units
B 14 units
C 237 units
D \(\sqrt{237}\) units

Verified Solution

Answer & Solution

Correct Answer

(A) \(\sqrt{14}\) units

Step-by-step Solution

Detailed explanation

Let \(\mathrm{P}=(2,-1,5)\) and co-ordinates of any point on the given line be
\(\mathrm{Q} \equiv(10 \lambda+11,-4 \lambda-2,-11 \lambda-13) \)
\( \text { d.r. of } \mathrm{PQ} \text { are }(10 \lambda+9,-4 \lambda-1,-11 \lambda-13) \)
\( \text { d.r. of given line are }(10,-4,-11) \)
\( \therefore(10 \lambda+9)(10)+(-4 \lambda-1)(-4)+(-11 \lambda-13)(-\) \(11)=0 \)
\( \therefore 100 \lambda+90+16 \lambda+4+121 \lambda+143=0 \Rightarrow \lambda=-1 \)
\( \therefore \mathrm{Q} \equiv(1,2,3) \text { and } \mathrm{d}(\mathrm{PQ})=\sqrt{1^2+3^2+2^2}\) \(=\sqrt{14} \text { units }\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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