Write the equation of line passing through
the point with position vector i +2j+3k
and in the direction 3i+2j - 2k in vector
form.
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Answer:
The vector equation of a line passing through a point with position vector
a
and parallel to
b
is
r
=
a
+λ
b
Given the line passes through 2
i
^
−3
j
^
+4
k
^
So,
a
=2
i
^
−3
j
^
+4
k
^
Finding normal of plane
r
.(3
i
^
+4
j
^
−5
k
^
)=7
Comparing with
r
.
n
=d we have
n
=3
i
^
+4
j
^
−5
k
^
Since line is perpendicular to the plane, the line will be parallel to the normal of the plane.
∴
b
=
n
=3
i
^
+4
j
^
−5
k
^
Hence,
r
=(2
i
^
−3
j
^
+4
k
^
)+λ(3
i
^
+4
j
^
−5
k
^
)
∴ Vector equation of line is
r
=(2
i
^
−3
j
^
+4
k
^
)+λ(3
i
^
+4
j
^
−5
k
^
)
Cartesian form is x
i
^
+y
j
^
+z
k
^
=(2+λ)
i
^
+(−3+4λ)
j
^
+(4−5λ)
k
^
⇒x=2+λ,y=−3+4λ,z=4−5λ
⇒λ=x−2=
4
y+3
=
−5
z−4
Which is the required cartesian equation of the line
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