The equation of line of intersection of two planes 4x−4y−z+11=x+2y−z−1=0 in symmetric form is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$\frac{x}{2} = \frac{y - 2}{1} = \frac{z - 3}{4}$
B
$\frac{x - 2}{2} = \frac{y - 2}{1} = \frac{z}{4}$
C
$\frac{x - 2}{2} = \frac{y}{1} = \frac{z - 3}{4}$
D
$\frac{x - 2}{2} = \frac{y}{- 1} = \frac{z + 3}{4}$

Solution:

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

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

$y = 2, z = 3$

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

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

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

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

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

Solution:

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