If the lines x-1-3-y-22 k-z-32- x-13 k-5-y-1-6-z 5 are at right angle- then k-

Two #160;lines #160;x x1a1=y y1b1=z z1c1 #160;and #160;x x2a2=y y2b2=z z2c2 #160;are #160;at #160;right #160;angle #160;if #160;a1 #160;a2 ...

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Byju's Answer

Standard X

Mathematics

Equation of a Line Parallel to a Straight Line

If the lines ...

Question

If the lines $\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}, \frac{x - 1}{3 k} = \frac{5 - y}{- 1} = \frac{6 - z}{5}$ are at right angle, then k = _______________ .

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Solution

$Two lines \frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} and \frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}} are at right angle if a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 ∴ \frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} and \frac{x - 1}{3 k} = \frac{5 - y}{- 1} = \frac{6 - z}{5} or \frac{x - 1}{3 k} = \frac{y - 5}{1} = \frac{z - 6}{- 5} are at right angle if (- 3) (3 k) + 2 k (+ 1) + 2 (- 5) = 0 i . e . - 9 k + 2 k - 10 = 0 i . e . - 7 k = 10 i . e . k = \frac{- 10}{7}$

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If the lines x-1-3=y-22k=z-32, x-13k=5-y-1=6-z5 are at right angle, then k = _______________ .

Two lines x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 are at right angle if a1 a2+b1 b2+c1 c2=0∴ x-1-3=y-22k=z-32 and x-13k=5-y-1=6-z5 or x-13k=y-51=z-6-5are at right angleif -3 3k+2k(+1)+2-5=0i.e. -9k+2k-10=0i.e. -7k=10 i.e. k=-107

If the lines $\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}, \frac{x - 1}{3 k} = \frac{5 - y}{- 1} = \frac{6 - z}{5}$ are at right angle, then k = _______________ .