Question

A line passing through A(1, 2, 3) and having direction ratios (3, 4, 5) meets a plane x + 2y – 3z = 5 at B, then distance AB is equal to-

• A. $\frac { 9 } { 4 }$
• B. $\frac { 11 } { 4 }$
• C. $\frac { 13 } { 4 }$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

Equation of the line is = $\frac { y - 2 } { 4 }$=

i.e., = = = r (say)

\ 1 + $\frac { 3 r } { 5 \sqrt { 2 } }$ + 4 + – 9 – = 5

̃ – = 9 ̃ r = $\frac { 45 \sqrt { 2 } } { 4 }$.

̃ A(1, 2, 3) and B $\left( \frac { 31 } { 4 } , 11 , \frac { 57 } { 4 } \right)$

\ |AB| = $\sqrt { \left( \frac { 31 } { 4 } - 1 \right) ^ { 2 } + ( 11 - 2 ) ^ { 2 } + \left( \frac { 57 } { 4 } - 3 \right) ^ { 2 } }$