The shortest distance between the lines $x - y = 0 = 2 x + z$ and $x + y - 2 = 0 = 3 x - y + z - 1$ is

\frac{1}{\sqrt{11}}

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\frac{1}{2 \sqrt{3}}

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\frac{1}{2}

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1

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Solution

The correct option is D $\frac{1}{\sqrt{11}}$
$L_{1} : x - y = 0 = 2 x + z$ and
$L_{2} : x + y - 2 = 0 = 3 x - y + z - 1$
It can also be write as
$L_{1} : \frac{x}{1} = \frac{y}{1} = \frac{z}{- 2}$
$L_{2} : \frac{x}{- 1} = \frac{y - 2}{1} = \frac{z - 3}{4}$

$A (0, 0, 0)$ be the point on $L_{1}$ and $\to B = (^i +^j - 2^k)$ is a direction of $L_{1}$
$C (0, 2, 3)$ be the point on $L_{2}$ and $\to D = (-^i +^j + 4^k)$ is a direction of $L_{2}$
$\to B \times \to D = 6^i - 2^j + 2^k$

$∣ ∣ \to B \times \to D ∣ ∣ = \sqrt{6^{2} + 2^{2} + 2^{2}} = 2 \sqrt{11}$

$\to C - \to A = 2^j + 3^k$

$(\to B \times \to D) . (\to C - \to A) = (6^i - 2^j + 2^k) \cdot (2^j + 3^k) = 2$

The shortest distance between line $L_{1}$ and $L_{2}$ can bee found using the following formula.
Distance $= ∣ ∣ ∣ ∣ \frac{\to B \times \to D}{| \to B \times \to D |} . (\to C - \to A) ∣ ∣ ∣ ∣$
Distance $= \frac{1}{\sqrt{11}}$

Hence, option A.

Suggest Corrections

Similar questions

\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1}

x + y + z + 1 = 0, 2 x - y + z + 3 = 0

\frac{x - 1}{2} = \frac{y + 1}{0} = \frac{z}{1}

\frac{x - 2}{2} = \frac{y + 1}{0} = \frac{z}{1}

\frac{x - 1}{α} = \frac{y + 1}{- 1} = \frac{z}{1}, (α \neq - 1)

x + y + z + 1 = 0 = 2 x - y + z + 3

\frac{1}{\sqrt{3}},

α

\frac{x - 2}{- 1} = \frac{y - 5}{2} = \frac{z - 0}{3} a n d \frac{x - 0}{2} = \frac{y + 1}{- 1} = \frac{z - 1}{2}

\frac{x - 3}{- 1} = \frac{y - 4}{2} = \frac{z + 2}{1}, \frac{x - 1}{1} = \frac{y + 7}{3} = \frac{z + 2}{2}

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