$P, Q, R$ 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 $P Q$ meets the line $R S$ at the point.

3 ¯ i + 4 ¯ j + 3 ¯ ¯ ¯ k

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- 3 ¯ i + 4 ¯ j + 3 ¯ ¯ ¯ k

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- ¯ i + 4 ¯ j + ¯ ¯ ¯ k

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¯ i + ¯ j + ¯ ¯ ¯ k

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Solution

The correct option is A $- 3 ¯ i + 4 ¯ j + 3 ¯ ¯ ¯ k$
$\begin{matrix} P (3, - 4, 5), Q (0, 0, 4), R (- 4, 5, 1), S (- 3, 4, 3) L_{1} : \frac{x - 3}{3} = \frac{y + 4}{- 4} = \frac{z - 5}{1} (3 r + 3, - 4 r - 4, r + 5) L_{2} : \frac{x + 4}{- 1} = \frac{y - 5}{1} = \frac{x - 1}{- 2} \frac{3 r + 7}{- 1} = \frac{- 4 r - 9}{1} = \frac{r + 4}{- 2} 2 (3 r + 7) = r + 4 5 r + 10 = 0 r = - 2 (- 3, 4, 3) - 3^i + 4^j + 3^k H e n c e, c o r r e c t o p t i o n i s (B) \end{matrix}$

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Similar questions

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

Q. Show that the points whose position vectors are as given below are collinear:
(i)

2 \hat{i} + \hat{j} - \hat{k}, 3 \hat{i} - 2 \hat{j} + \hat{k} and \hat{i} + 4 \hat{j} - 3 \hat{k}

(ii)

3 \hat{i} - 2 \hat{j} + 4 \hat{k}, \hat{i} + \hat{j} + \hat{k} and - \hat{i} + 4 \hat{j} - 2 \hat{k}

Q. The three points whose position vectors are

¯ i + 2 ¯ j + 3 ¯ ¯¯ ¯ k,

3 ¯ i + 4 ¯ j + 7 ¯ ¯¯ ¯ k,

and

- 3 ¯ i - 2 ¯ j - 5 ¯ ¯ ¯ k

Q. The triangle formed by the three points whose position vectors are

2 i + 4 j - k, 4 i + 5 j + k

and

3 i + 6 j - 3 k

, is

Q. Prove that the points having position vectors

\hat{i} + 2 \hat{j} + 3 \hat{k}, 3 \hat{i} + 4 \hat{j} + 7 \hat{k}, - 3 \hat{i} - 2 \hat{i} - 5 \hat{k}

are collinear.

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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 A −3¯i+4¯j+3¯¯¯kP(3,−4,5),Q(0,0,4),R(−4,5,1),S(−3,4,3)L1:x−33=y+4−4=z−51(3r+3,−4r−4,r+5)L2:x+4−1=y−51=x−1−23r+7−1=−4r−91=r+4−22(3r+7)=r+45r+10=0r=−2(−3,4,3)−3^i+4^j+3^kHence,correctoptionis(B)

$P, Q, R$ 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 $P Q$ meets the line $R S$ at the point.