The equation of line of intersection of two planes 4x−4y−z+11=x+2y−z−1=0 in symmetric form is

# The equation of line of intersection of two planes $4x-4y-z+11=x+2y-z-1=0$ in symmetric form is

1. A

$\frac{x}{2}=\frac{y-2}{1}=\frac{z-3}{4}$

2. B

$\frac{x-2}{2}=\frac{y-2}{1}=\frac{z}{4}$

3. C

$\frac{x-2}{2}=\frac{y}{1}=\frac{z-3}{4}$

4. D

$\frac{x-2}{2}=\frac{y}{-1}=\frac{z+3}{4}$

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### Solution:

The given planes are $4x-4y-z+11=x+2y-z-1=0$

To get a point on the line, substitute $x=0$ in above equations and then solve for the other coordinates

Hence, a point on the line is $\left(0,2,3\right)$

The direction ratios are proportional to the components of the vector $\left|\begin{array}{ccc}i& j& k\\ 4& -4& -1\\ 1& 2& -1\end{array}\right|=i\left(6\right)-j\left(-3\right)+k\left(12\right)$

Hence, the direction ratios of the required line are in proportion with $⟨2,1,4⟩$

Therefore, the equation of the line is $\frac{x}{2}=\frac{y-2}{1}=\frac{z-3}{4}$

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