Consider the L1: x+13=y+21=z+12, L2: x−21=y+22=z−33 lines Which of the following is not Correct?

The vector perpendicular to L1 and L2 is |i^ j^ k^ 3 1 21 2 3|=−i^ −7 j^ +5k^∴ Unit vector (n^) is =−i^ −7 j^ +5k^53.The Points on L1 and L2 are A(−1,−2,−1) and C(2,−2,3)The shortest distance is A¯C¯.n^=(3i^+4k^).−i^ −7 j^ +5k^53=1753The d.r.’s of the normal are -1,-7,5∴ The plane is −(x+1)−7(y+2)+5(z+1)=0⇒x+7y−5z+10=0The distance of (1,1,1) is 13/75=1353

Consider the L1: x+13=y+21=z+12, L2: x−21=y+22=z−33 lines Which of the following is not Correct?

The vector perpendicular to L1 and L2 is |i^ j^ k^ 3 1 21 2 3|=−i^ −7 j^ +5k^∴ Unit vector (n^) is =−i^ −7 j^ +5k^53.The Points on L1 and L2 are A(−1,−2,−1) and C(2,−2,3)The shortest distance is A¯C¯.n^=(3i^+4k^).−i^ −7 j^ +5k^53=1753The d.r.’s of the normal are -1,-7,5∴ The plane is −(x+1)−7(y+2)+5(z+1)=0⇒x+7y−5z+10=0The distance of (1,1,1) is 13/75=1353

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Consider the $L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}, L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$ lines Which of the following is not Correct?

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The distance of the point (1,1,1) from the plane passing through (-1,-2,-1) and whose normal is perpendicular to both $L_{1}, L_{2} i s \frac{13}{7 \sqrt{3}}$

The shortest distance between $L_{1} a n d L_{2} i s \frac{17}{5 \sqrt{3}}$

The distance of the point (1,1,1) from the plane passing through (-1,-2,-1) and whose normal is perpendicular to both $L_{1}, L_{2} i s \frac{13}{5 \sqrt{3}}$

The unit vector perpendicular to both $L_{1} & L_{2} i s \frac{- \hat{i} - 7 \hat{j} + 5 \hat{k}}{5 \sqrt{3}}$

answer is D.

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Detailed Solution

$T h e v e c t o r p e r p e n d i c u l a r t o L_{1} a n d L_{2} i s | \begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 3 1 2 \\ 1 2 3 \end{array} | = - \hat{i} - 7 \hat{j} + 5 \hat{k}$

$∴ U n i t v e c t o r (\hat{n}) i s = \frac{- \hat{i} - 7 \hat{j} + 5 \hat{k}}{5 \sqrt{3}} .$

The Points on $L_{1} a n d L_{2} a r e A (- 1, - 2, - 1) a n d C (2, - 2, 3)$

The shortest distance is $\bar{A} \bar{C} . \hat{n} = (3 \hat{i} + 4 \hat{k}) . \frac{- \hat{i} - 7 \hat{j} + 5 \hat{k}}{5 \sqrt{3}} = \frac{17}{5 \sqrt{3}}$
The d.r.’s of the normal are -1,-7,5
$\begin{array}{l} ∴ T h e p l a n e i s - (x + 1) - 7 (y + 2) + 5 (z + 1) = 0 \\ \Rightarrow x + 7 y - 5 z + 10 = 0 \\ T h e d i s \tan c e o f (1, 1, 1) i s 13 / \sqrt{75} \\ = \frac{13}{5 \sqrt{3}} \end{array}$