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MHT CET · Maths · Three Dimensional Geometry

The length of the perpendicular from the point \((0,2,3)\) on the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\)

  1. A \(\sqrt{15}\) units
  2. B \(\sqrt{21}\) units
  3. C \(\sqrt{33}\) units
  4. D \(\sqrt{11}\) units
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{21}\) units

Step-by-step Solution

Detailed explanation

Any point on the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda\) can be taken as
\(
(5 \lambda-3,2 \lambda+1,3 \lambda-4)
\)
\(\text {for foot of perpendicular } \lambda=\) \(\frac{a\left(\alpha-x_1\right)+b\left(\beta-y_1\right)+c\left(\gamma-z_1\right)}{a^2+b^2+c^2} \)
\( =\frac{5(0+3)+2(2-1)+3(3+4)}{5^2+2^2+3^2} \)
\( \text {i.e., } \lambda=1 \)
\( \Rightarrow \text {foot of perpendicular }(2,3,-1) \)
\( \Rightarrow \text {Length of the perpendicular }=\) \(\sqrt{(0-2)^2+(2-3)^2+(3+1)^2} \)
\( =\sqrt{21}\)