CUET UG Mathematics Applied Straight Line Previous Year Questions (PYQs)
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The point of intersection of the lines
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$
and
$\frac{x-4}{5}=\frac{y-1}{2}=z$
is:
1
2
3
4
Solution
First line
Let parameter $t$
$x=1+2t$
$y=2+3t$
$z=3+4t$
Second line
Let parameter $s$
$\frac{x-4}{5}=\frac{y-1}{2}=z=s$
So
$x=4+5s$
$y=1+2s$
$z=s$
For intersection
$3+4t=s$
Substitute in $x$
$1+2t=4+5(3+4t)$
$1+2t=19+20t$
$-18=18t$
$t=-1$
Then
$s=3+4(-1)=-1$
Point
$x=1+2(-1)=-1$
$y=2+3(-1)=-1$
$z=3+4(-1)=-1$
The distance between the point $(3,4,5)$ and the point where the line
$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$
meets the plane
$x+y+z=17$
is:
1
2
3
4
Solution
Line in parametric form
Let parameter $t$
$x=3+t$
$y=4+2t$
$z=5+2t$
Substitute in plane
$x+y+z=17$
$(3+t)+(4+2t)+(5+2t)=17$
$12+5t=17$
$t=1$
Point of intersection
$x=4$
$y=6$
$z=7$
Distance from $(3,4,5)$
$d=\sqrt{(4-3)^2+(6-4)^2+(7-5)^2}$
$d=\sqrt{1+4+4}$
$d=3$
CUET UG Mathematics Applied
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CUET UG Mathematics Applied
Online Test Series, Information About Examination,
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and More.
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