Let P be the point (10,-2,-1) and Q be the foot of the perpendicular drawn from the point R(1,7,6) on the line passing through the points (2,-5,11) and (-6,7,-5) . Then the length of the line segment P Q is equal to \underline\hspace2cm .

3D Geometry PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q55 with Solution

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Correct answer: 13

The line passes through points $A(2,-5,11), \quad B(-6,7,-5).$

Its direction vector is $\vec d = B-A = (-6-2,\; 7-(-5),\; -5-11) = (-8,12,-16).$

We can simplify: $\vec d = (-2,3,-4).$

So a general point on the line is $Q = A + t\vec d = (2,-5,11) + t(-2,3,-4).$

Hence $Q=(2-2t,\,-5+3t,\,11-4t).$

Since $Q$ is the foot of the perpendicular from $R$ to the line, we must have $\overrightarrow{RQ} \perp \vec d.$

Now $\overrightarrow{RQ}=Q-R=(2-2t-1,\,-5+3t-7,\,11-4t-6)=(1-2t,\,-12+3t,\,5-4t).$

Perpendicularity gives $\overrightarrow{RQ}\cdot \vec d=0.$

So, $(1-2t)(-2)+(-12+3t)(3)+(5-4t)(-4)=0.$

Compute: $-2+4t-36+9t-20+16t=0$ $29t-58=0$ $t=2.$

Therefore,

-5+3\cdot 2, 11-4\cdot 2)=(-2,1,3).$$ --- 3. **Find distance** $PQ$ Given $$P=(10,-2,-1), \quad Q=(-2,1,3).$$ Then $$\overrightarrow{PQ}=Q-P=(-2-10,\,1-(-2),\,3-(-1))=(-12,3,4).$$ Hence $$PQ=\sqrt{(-12)^2+3^2+4^2}$$ $$=\sqrt{144+9+16}$$ $$=\sqrt{169}$$ $$=13.$$ --- 4. **Final Answer** $$\boxed{13}$$ The derived answer matches the stored correct answer.