First slide

Straight lines in 3D

Question

The symmetric form of the line of intersection of planes $3 x + 2 y + z - 5 = 0$ and $x + y - 2 z = 0$

Moderate

A

$\frac{x - 5}{5} = \frac{y + 5}{- 7} = \frac{z}{- 1}$
B

$\frac{x - 5}{5} = \frac{y + 5}{7} = \frac{z}{- 1}$
C

$\frac{x - 5}{5} = \frac{y - 5}{- 7} = \frac{z}{- 1}$
D

$\frac{x - 5}{5} = \frac{y - 5}{7} = \frac{z}{1}$

Solution

The given planes are $x + y - 2 z = 0$ and $3 x + 2 y + z - 5 = 0$ $x = 5, y = - 5$
Plug in $z = 0$ to get a point on the line.
$x + y = 0, 3 x + 2 y - 5 = 0$
solve the above two equations, we get $x = 5, y = - 5$
A point on the line is $(5, - 5, 0)$
Vector along the line is cross product of two normal vectors to the planes
it gives $n_{1} \times n_{2} = |\begin{matrix} i & j & k \\ 1 & 1 & - 2 \\ 3 & 2 & 1 \end{matrix}| = i (5) - j (7) + k (- 1)$
Direction ratios of a line of intersection of two planes are proportional to $⟨ 5, - 7, - 1 ⟩$
Therefore, the equation of the required line is $\frac{x - 5}{5} = \frac{y + 5}{- 7} = \frac{z}{- 1}$

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