Let point P lies on line r-5-7-2k-s3-k- and point Q lies on the line r-3-3-6k-t-3-2-4k- if PQ is parallel to vector 2-7-5k- then -PQ-2-

Given : nbsp; P lies on line r=5î+7ĵ 2k̂+s(3î ĵ+k̂) and nbsp; point Q lies on the line r= 3î+3ĵ+6k̂+t( 3î+2ĵ+4k̂). PQ=OQ OP= 8î 4ĵ+8k̂+t( 3î+2ĵ+4k̂) s(3î ĵ+ ...

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Standard XII

Mathematics

Slope of a Line Connecting Two Points

Let point P l...

Question

Let point $P$ lies on line $\to r = 5^i + 7^j - 2^k + s (3^i -^j +^k)$ and point $Q$ lies on the line $\to r = - 3^i + 3^j + 6^k + t (- 3^i + 2^j + 4^k) .$ if $- - \to P Q$ is parallel to vector $2^i + 7^j - 5^k,$ then $| - - \to P Q |^{2} =$

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Solution

Given : $P$ lies on line $\to r = 5^i + 7^j - 2^k + s (3^i -^j +^k)$ and point $Q$ lies on the line $\to r = - 3^i + 3^j + 6^k + t (- 3^i + 2^j + 4^k) .$
$- - \to P Q = - - \to O Q - - - \to O P = - 8^i - 4^j + 8^k + t (- 3^i + 2^j + 4^k) - s (3^i -^j +^k)$
As, $- - \to P Q$ is parallel to $2^i + 7^j - 5^k,$ Let $- - \to P Q = λ (2^i + 7^j - 5^k)$
$∴ 2 λ = - 8 - 3 t - 3 s \dots (i) 7 λ = - 4 + 2 t + s \dots (i i) - 5 λ = 8 + 4 t - s \dots (i i i)$
Solving, $(i), (i i), (i i i) :$ we get
$λ = t = s = - 1$
So, $- - \to P Q = - 2^i - 7^j + 5^k$
$∴ | - - \to P Q |^{2} = 78$

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Similar questions

Q. If

\to a = 2^i +^j +^k, \to b = 3^i - 4^j + 2^k, \to c =^i - 2^j + 2^k

then the projection of

\to a + \to b

\to c

\to a = 3^i +^j - 2^k; \to b =^i + λ^j - 3^k

\to a ⊥ \to b

, find value of

λ

Q. If

\to a

and

\to b

are vectors such that

| \to a + \to b | = \sqrt{29}

and

\to a \times (2^i + 3^j + 4^k) = (2^i + 3^j + 4^k) \times \to b

, then a possible value of

(\to a + \to b) . (- 7^i + 2^j + 3^k)

Q. Find the shortest distance between the lines

¯ ¯ ¯ r = (4^i -^j) + λ (^i + 2^j - 3^k)

and

¯ ¯ ¯ r = (^i -^n j + 2^k) + μ (^i + 4^j - 5^k)

, where

λ

and

μ

are parameters.

Q. Find

λ

if the vectors

\to a =^i + 3^j +^k \to b = 2^i -^j -^k \to c = λ^i + 7^j + 3^k

are coplanar.

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Slope of a Line Connecting Two Points

Standard XII Mathematics

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Let point P lies on line →r=5^i+7^j−2^k+s(3^i−^j+^k) and point Q lies on the line →r=−3^i+3^j+6^k+t(−3^i+2^j+4^k). if −−→PQ is parallel to vector 2^i+7^j−5^k, then |−−→PQ|2=

Let point $P$ lies on line $\to r = 5^i + 7^j - 2^k + s (3^i -^j +^k)$ and point $Q$ lies on the line $\to r = - 3^i + 3^j + 6^k + t (- 3^i + 2^j + 4^k) .$ if $- - \to P Q$ is parallel to vector $2^i + 7^j - 5^k,$ then $| - - \to P Q |^{2} =$