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

Equation of PQ:x/3=y/ 4=z 4/1 a point A on PQ=(3λ, 4λ,λ+4) Equation of RSx+4/1=y 5/ 1=z 1/2=m a point on RS, B=(m 4,m+5,2m+1) to intersect A = B⇒λ= 1,m=1 ∴Point ...

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Standard XII

Mathematics

Projection of a Line Segment on a Line

P,Q,R and S a...

Question

P,Q,R and S are four points with the position vectors, $3 \to i - 4 \to j + 5 \to k, 4 \to k, - 4 \to i + 5 \to j + \to k and - 3 \to i + 4 \to j + 3 \to k$ 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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$- \to i + 4 \to j + \to k$

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$\to i + \to j + \to k$

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Solution

The correct option is B
$- 3 \to i + 4 \to j + 3 \to k$

$Equation of PQ : \frac{x}{3} = \frac{y}{- 4} = \frac{z - 4}{1} a point A on PQ = (3 λ, - 4 λ, λ + 4) Equation of RS \frac{x + 4}{1} = \frac{y - 5}{- 1} = \frac{z - 1}{2} = m a point on RS, B = (m - 4, m + 5, 2 m + 1) to intersect A = B \Rightarrow λ = - 1, m = 1 ∴ Point is (- 3, 4, 3)$

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Similar questions

Q. 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

Q. Show that the points whose position vectors

4 \to i + 5 \to j + \to k, 3 \to i + 9 \to j + 4 \to k

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

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Q. If the position vectors of

A, B

and

C

are respectively

2 \to i - \to j + \to k, \to i - 3 \to j - 5 \to k

and

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, then

{cos}^{2} A

is equal to

Q. The ratio in which the plane

\to r \cdot (\to i - 2 \to j + 3 \to k) = 17

divides the line joining the points

- 2 \to i + 4 \to j + 7 \to k

and

3 \to i - 5 \to j + 8 \to k

is-

Q. Show that the vectors

2 \to i - \to j + \to 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+→k and −3→i+4→j+3→k respectively. Then the line PQ meets the line RS at the point

The correct option is B −3→i+4→j+3→k Equation of PQ:x3=y−4=z−41a point A on PQ=(3λ,−4λ,λ+4)Equation of RSx+41=y−5−1=z−12=ma point on RS, B=(m−4,m+5,2m+1)to intersect A = B⇒λ=−1,m=1∴Point is (−3,4,3)

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