Find the point of intersection of the lines L- x-31-y-2-1-z-1-1 and M- x2-y-33 - z-1-

PI of Intersection of x 3/c = y+2/ 1 = z+1/ 1 x/2 = y+3/3 = z 1 #160; x 3/1 = x/2⇒ x = 6 y+3/3 = y+2/ 1⇒ y = 9/4 3+1/ 1 = z 1⇒ 3 = 0 ...

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Standard XII

Mathematics

Cross Product of Two Vectors

Find the poin...

Question

Find the point of intersection of the lines L: $\frac{x - 3}{1} = \frac{y + 2}{- 1} = \frac{z + 1}{- 1}$ and M: $\frac{x}{2} = \frac{y + 3}{3}$ = $z - 1$ .

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Solution

PI of Intersection of $\frac{x - 3}{c} = \frac{y + 2}{- 1} = \frac{z + 1}{- 1}$
$\frac{x}{2} = \frac{y + 3}{3} = z - 1$
$\frac{x - 3}{1} = \frac{x}{2} \Rightarrow x = 6$
$\frac{y + 3}{3} = \frac{y + 2}{- 1} \Rightarrow y = - 9 / 4$
$\frac{3 + 1}{- 1} = z - 1 \Rightarrow 3 = 0$
$∴ P . I (6, \frac{- 9}{4}, 0)$

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Find the point of intersection of the lines L: x−31=y+2−1=z+1−1 and M: x2=y+33 = z−1.

PI of Intersection of x−3c=y+2−1=z+1−1x2=y+33=z−1 x−31=x2⇒x=6y+33=y+2−1⇒y=−9/43+1−1=z−1⇒3=0∴P.I(6,−94,0)

Find the point of intersection of the lines L: $\frac{x - 3}{1} = \frac{y + 2}{- 1} = \frac{z + 1}{- 1}$ and M: $\frac{x}{2} = \frac{y + 3}{3}$ = $z - 1$ .