Solveeit Logo
Login

Question

Mathematics Question on Three Dimensional Geometry

If the three planes x = 5, 2x - 5ay + 3z - 2 = 0 and 3bx + y - 3z = 0 contain a common line, then (a, b) is equal to

A

(815,−15)\left(\frac{8}{15}, - \frac{1}{5}\right)

B

(15,−815)\left( \frac{1}{5}, -\frac{8}{15}\right)

C

(−815,15)\left( - \frac{8}{15},\frac{1}{5}\right)

D

(−15,815)\left(- \frac{1}{5}, \frac{8}{15}\right)

Answer

(15,−815)\left( \frac{1}{5}, -\frac{8}{15}\right)

Explanation

Solution

Let the direction ratios of the common line be l, m and n. ∴ l×1+m×0+n×0=0 ⇒ l=0\therefore \, l \times 1 + m \times 0 + n \times 0 = 0 \, \Rightarrow \, l = 0 ....(1) 2l−5ma+3n=0 ⇒ 5ma−3n=02l - 5ma + 3n = 0 \, \Rightarrow \, 5ma - 3n = 0 ....(2) 3lb+m−3n=0 ⇒ m−3n=03lb + m - 3n = 0 \, \Rightarrow \, m - 3n = 0 ....(3) Subtracting (3) from (1), we get m(5a−1)=0m(5a - 1) = 0 Now, value of m can not be zero because if m = 0 then n = 0 ⇒ l=m=n=0\Rightarrow \, l = m = n = 0 which is not possible. Hence, 5a - 1 = 0 ⇒ a=15\Rightarrow \, a = \frac{1}{5}