The vector of the straight line {x-2}/1=y/{-3}={1-z}/2\is

Correct Answer is : →r=2⋀i+⋀k+t(⋀i−3⋀j−2⋀k)

r^{\to}=2i^{⋀}+k^{⋀}+t(i^{⋀}+3j^{⋀}+2k^{⋀})

r^{\to}=2i^{⋀}-k^{⋀}+t(i^{⋀}-3j^{⋀}-2k^{⋀})

r^{\to}=2i^{⋀}+k^{⋀}+t(i^{⋀}-3j^{⋀}+2k^{⋀})

r^{\to}=2i^{⋀}+j^{⋀}+t(i^{⋀}-3j^{⋀}-2k^{⋀})

r^{\to}=2i^{⋀}+k^{⋀}+t(i^{⋀}-3j^{⋀}-2k^{⋀})

KEAM 2016 Math Paper

Question : 79 of 120

Marks: +1, -0

The vector of the straight line

x−2

−3

1−z

‌is

→
r
=2
⋀
i
+
⋀
k
+t(
⋀
i
+3
⋀
j
+2
⋀
k
)
→
r
=2
⋀
i
−
⋀
k
+t(
⋀
i
−3
⋀
j
−2
⋀
k
)
→
r
=2
⋀
i
+
⋀
k
+t(
⋀
i
−3
⋀
j
+2
⋀
k
)
→
r
=2
⋀
i
+
⋀
j
+t(
⋀
i
−3
⋀
j
−2
⋀
k
)
→
r
=2
⋀
i
+
⋀
k
+t(
⋀
i
−3
⋀
j
−2
⋀
k
)

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