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KEAM 2010 Math Paper

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Question : 13 of 120

Marks: +1, -0

If the plane

3x+y+2z+6=0

is parallel to the line

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

then the value of

3a+3b

is

$\frac{1}{2}$
$\frac{3}{2}$
3
4
$\frac{5}{2}$

Solution:

Comparing the given equation of plane

3x+y+2z+6=0

with

lx+my+nz+d=0

\Rightarrow\; l=3,m=1,n=2

Also, comparing given equation of line

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

with

\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1},

we get

a_1 = \frac{2b}{3}, b_1 = -1, c_1 = a

For parallel line

l a_1 + m b_1 + n c_1 = 0

\Rightarrow\; 3 \cdot \frac{2b}{3} + 1 \cdot (-1) + 2a = 0

\Rightarrow\; 2a+2b=1

\Rightarrow\; a+b = \frac{1}{2}

\Rightarrow\; 3a+3b = \frac{3}{2}

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