The distance of the point - i+2 j+6 k from the straight line that passes through the point 2 i+3 j-4 k and is parallel to the vector 6 i+3 j-4 k is
JEE Main 2012 — Mathematics Vectors & 3D Geometry
2012mcqmedium
The distance of the point −i^+2j^+6k^ from the straight line that passes through the point 2i^+3j^−4k^ and is parallel to the vector 6i^+3j^−4k^ is
Official previous-year question
Held on 26 May 2012 · Verified 6 Jul 2026.
Options
A
9
B
8
C
7
D
10
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Solution
Point is (−1,2,6) Line passes through the point (2,3,−4) parallel to vector whose direction ratios is 6,3,−4. Equation is 6x−2=3y−3=−4z+4=λ Any point on this line is given by x=6λ+2,y=3λ+3,z=−4λ−4 Now, d. Rs of line passing through (−1,2,6) and ⊥ to this line is {(x+1),(y−2),(z−6)} So, 6(x+1)+3(y−2)−4(z−6)=0⇒6x+3y−4z+24=0 Now, 6(6λ+2)+3(3λ+3)+4(4λ+4)+24=0⇒61λ+61=0⇒λ=−1 So, x=−4,y=0,z=0 Now, distance between (−1,2,6) and (−4,0,0) is 9+4+36=49=7
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