If the plane 3 x+y+2 z+6=0 is parallel to the line (3 x-1)/(2 b)=3-y=(z-1)/(a) , then the value of 3 a+3 b is

Correct answer: B. (3)/(2) Given plane 3 x+y+2 z+6=0 and line (x-(1)/(3))/((2 b)/(3))=(y-3)/(-1)=(z-1)/(a) Since, plane is parallel to line, then aligned & 3((2 b)/(3))+(1)(-1)+2(a)=0 & 2 b-1+2 a=0 & a+b=(1)/(2) & aligned Now, 3 a+3 b=(3)/(2)

BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Mathematics
Chapter: Three Dimensional Geometry

BITSAT 2022 · Mathematics · Three Dimensional Geometry

If the plane 3 x+y+2 z+6=0 is parallel to the line (3 x-1)/(2...

If the plane

3 x+y+2 z+6=0

is parallel to the line

\frac{3 x-1}{2 b}=3-y=\frac{z-1}{a}

, then the value of

3 a+3 b

Options

A $\frac{1}{2}$
B $\frac{3}{2}$
C3
D4

Correct answer

B. $\frac{3}{2}$

Step-by-step solution

Given plane

3 x+y+2 z+6=0

and line

\frac{x-\frac{1}{3}}{\frac{2 b}{3}}=\frac{y-3}{-1}=\frac{z-1}{a}

Since, plane is parallel to line, then

\begin{aligned}& 3\left(\frac{2 b}{3}\right)+(1)(-1)+2(a)=0 \\ & 2 b-1+2 a=0 \\ & \Rightarrow \quad a+b=\frac{1}{2}\\ & \end{aligned}

Now, 3 a+3 b=\frac{3}{2}$

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