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Maths Question 13 – JEE-MAIN 2026

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The square of the distance of the point P(5, 6, 7) from the line x−22=y−53=z−24 is equal to:

The distance from a point to a line in 3D space can be found using vector operations.

● Video Walkthrough
Step 1: Identify Point and Line Parameters✦ Active

The given point is P0(5,6,7). The equation of the line is x−22=y−53=z−24. From this, we can identify a point A on the line and its direction vector b→:

A=(2,5,2) b→=(2,3,4)

Now, form the vector AP0→ from point A to P0:

AP0→=P0−A=(5−2,6−5,7−2)=(3,1,5)
Step 2: Calculate Cross Product and Magnitudes○ Expand

Compute the cross product AP0→×b→:

AP0→×b→=|ijk315234|=i(1⋅4−5⋅3)−j(3⋅4−5⋅2)+k(3⋅3−1⋅2) =i(4−15)−j(12−10)+k(9−2)=−11i−2j+7k

Now, calculate the square of the magnitude of this cross product and the square of the magnitude of the direction vector b→:

|AP0→×b→|2=(−11)2+(−2)2+(7)2=121+4+49=174 |b→|2=(2)2+(3)2+(4)2=4+9+16=29
💡 Teacher's Secret Hint

Ensure careful calculation of the determinant for the cross product to avoid sign errors.

Step 3: Compute the Square of the Distance○ Expand

The square of the distance d2 from the point to the line is given by the formula:

d2=|AP0→×b→|2|b→|2 d2=17429 d2=6

Thus, the square of the distance is 6.

💡 Teacher's Secret Hint

Double-check the final division to ensure accuracy.

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