P-Q-R and S are four points with the position vectors 3i- 4j- - 5k- 4k- -4i- - 5j- - k- and -3i- - 4j- - 3k- respectively- Then the line PQ meets the line RS at the point

The correct option is B 3i⃗ + 4j⃗ + 3k⃗ PQ=4k̂+λ[ ̂3̂î+4ĵ k̂]= 3λî+4λĵ+(4 λ) k̂) #160; RS =( 3i+4j+3k)+μ ( î+ĵ 2k̂)=( 3 μ)î+(4+μ)ĵ+(3 2μ)k̂ Fo ...

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Standard XII

Mathematics

Reflection of a Point about a Line

P,Q,R and S a...

Question

P,Q,R and S are four points with the position vectors $3 \to i - 4 \to j + 5 \to k, 4 \to k, - 4 \to i + 5 \to j + \to k a n d - 3 \to i + 4 \to j + 3 \to k$ respectively. Then the line PQ meets the line RS at the point

3 \to i + 4 \to j + 3 \to k

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- 3 \to i + 4 \to j + 3 \to k

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- \to i + 4 \to j + 3 \to k

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\to i + 4 \to j + \to k

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Solution

The correct option is B $- 3 \to i + 4 \to j + 3 \to k$
$- - \to P Q = 4^k + λ [^- 3 i + 4^j -^k] = - 3 λ^i + 4 λ^j + (4 - λ)^k)$

$- - \to R S = (- 3 i + 4 j + 3 k) + μ (-^i +^j - 2^k) = (- 3 - μ)^i + (4 + μ)^j + (3 - 2 μ)^k$

For pt. of intersection
$∴ - 3 λ = - 3 - μ$
$\Rightarrow 3 λ = 3 + μ$ ......(1)
And $4 λ = 4 + μ$ .......(2)

From eqn (1) and eqn (2)
$λ = 1$ and
$μ = 0$
Hence the point of intersection is $- 3^i + 4^j + 3^k$

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P,Q,R and S are four points with the position vectors 3→i−4→j+5→k,4→k,−4→i+5→j+→kand−3→i+4→j+3→k respectively. Then the line PQ meets the line RS at the point

P,Q,R and S are four points with the position vectors $3 \to i - 4 \to j + 5 \to k, 4 \to k, - 4 \to i + 5 \to j + \to k a n d - 3 \to i + 4 \to j + 3 \to k$ respectively. Then the line PQ meets the line RS at the point