A line with d-r-s 2-7-5 is drawn to intersect the lines x-53-y-7-1-z-21 and x-3-3-y-32-z-64 at P and Q respectively- Length of PQ is -

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Angle between Two Lines

A line with d...

Question

A line with d.r's $(2, 7, - 5)$ is drawn to intersect the lines $\frac{x - 5}{3} = \frac{y - 7}{- 1} = \frac{z + 2}{1}$ and $\frac{x + 3}{- 3} = \frac{y - 3}{2} = \frac{z - 6}{4}$ at $P$ and $Q$ respectively. Length of $P Q$ is :

\sqrt{78}

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\sqrt{54}

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\sqrt{74}

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Solution

The correct option is A $\sqrt{78}$
Any general point on the line can be taken as: $P = (5 + 3 t, 7 - t, - 2 + t); Q = (- 3 - 3 λ, 3 + 2 λ, 6 + 4 λ)$
$d.r's of PQ = (8 + 3 t + 3 λ, 4 - t - 2 λ, t - 4 λ - 8)$
But d.r's of $P Q$ is $(2, 7, - 5)$
Hence on comparing, we have $t = λ = - 1$
So, $P = (2, 8, - 3), Q = (0, 1, 2) \Rightarrow P Q = \sqrt{78}$

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Similar questions

A line with directiion ratios 2, 7, -5 is intercepted between the lines
$\frac{x - 5}{3} = \frac{y - 7}{- 1} = \frac{z + 2}{1} a n d \frac{x + 3}{- 3} = \frac{y - 3}{2} = \frac{z - 6}{4}$ . Find the length intercepted between the given lines.___

Q. Prove that the lines

\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}

and

\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}

are intersecting to each other. Find their point of intersection.

Q. The point of intersection of the lines

\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} a n d \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}

Q. Prove that the lines

[\frac{x + 3}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}] & [\frac{x + 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}]

not intersect at the point

[\frac{1}{2}, \frac{- 1}{2}, \frac{- 3}{2}]

Q. Show that the lines

\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}

and

\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}

intersect. Also find their point of intersection.

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A line with d.r's (2,7,−5) is drawn to intersect the lines x−53=y−7−1=z+21 and x+3−3=y−32=z−64 at P and Q respectively. Length of PQ is :

The correct option is A √78 Any general point on the line can be taken as:P=(5+3t,7−t,−2+t);Q=(−3−3λ,3+2λ,6+4λ) d.r's of PQ =(8+3t+3λ,4−t−2λ,t−4λ−8) But d.r's of PQ is (2,7,−5) Hence on comparing, we have t=λ=−1 So, P=(2,8,−3),Q=(0,1,2)⇒PQ=√78

A line with d.r's $(2, 7, - 5)$ is drawn to intersect the lines $\frac{x - 5}{3} = \frac{y - 7}{- 1} = \frac{z + 2}{1}$ and $\frac{x + 3}{- 3} = \frac{y - 3}{2} = \frac{z - 6}{4}$ at $P$ and $Q$ respectively. Length of $P Q$ is :