A line passes through 2-1-3 and is perpendicular to the lines r -k-2 -2 -k- andr -2 -3 k-2 -2 k- Obtain its equation in vector and cartesian form-

Given, lines are nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;x 1/2=y 1/ 2=z+1/1=λ ......(i) and nbsp; nbsp; x 2/1=y+1/2=z+3/2=μ......(ii) Suppose ...

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Byju's Answer

Standard XII

Mathematics

Distance between Two Parallel Planes

A line passes...

Question

A line passes through (2,-1,3) and is perpendicular to the lines
$\to r = (^i +^j -^k) + λ (2^i - 2^j +^k)$ and
$\to r = (2^i -^j - 3^k) + μ (^i + 2^j + 2^k)$ . Obtain its equation in vector and cartesian form.

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Solution

Given, lines are
$\frac{x - 1}{2} = \frac{y - 1}{- 2} = \frac{z + 1}{1} = λ . . . . . . (i)$

and $\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z + 3}{2} = μ . . . . . . (i i)$

$S u p p o s e e q u a t i o n o f l i n e p a s s e s t h r o u g h (2, - 1, 3) i s \frac{x - 2}{a} = \frac{y + 1}{b} = \frac{z - 3}{c} . . . . . . (i i i)$

$I f (i i i) i s ⊥ t o (i) a n d (i i)$

$∴ 2 a - 2 b + c = 0 a n d a + 2 b + 2 c = 0$

$\frac{a}{- 4 - 2} = \frac{b}{1 - 4} = \frac{c}{4 + 2}$

$\Rightarrow \frac{a}{- 6} = \frac{b}{- 3} = \frac{c}{6} \Rightarrow \frac{a}{2} = \frac{b}{1} = \frac{c}{- 2}$

$∴ from (iii), required cartesian form of line is \frac{x - 2}{2} = \frac{y + 1}{1} = \frac{z - 3}{- 2}$

$I t s v e c t o r f o r m i s \to r = (2^i -^j + 3^k) + s (2^i +^j - 2^k)$

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A line passes through (2,-1,3) and is perpendicular to the lines →r=(^i+^j−^k)+λ(2^i−2^j+^k) and →r=(2^i−^j−3^k)+μ(^i+2^j+2^k) . Obtain its equation in vector and cartesian form.

A line passes through (2,-1,3) and is perpendicular to the lines
$\to r = (^i +^j -^k) + λ (2^i - 2^j +^k)$ and
$\to r = (2^i -^j - 3^k) + μ (^i + 2^j + 2^k)$ . Obtain its equation in vector and cartesian form.