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Joint Entrance Examination
Mathematics
3D Geometry
Question 231
Question 231 of 521
Medium
MCQ
Let
O
O
O
be the origin and the position vector of
P
P
P
be
−
i
^
−
2
j
^
+
3
k
^
-\hat i-2\hat j+3\hat k
−
i
^
−
2
j
^
+
3
k
^
. If the position vectors of
A
,
B
,
C
A,B,C
A
,
B
,
C
are
−
2
i
^
+
j
^
−
3
k
^
-2\hat i+\hat j-3\hat k
−
2
i
^
+
j
^
−
3
k
^
,
2
i
^
+
4
j
^
−
2
k
^
2\hat i+4\hat j-2\hat k
2
i
^
+
4
j
^
−
2
k
^
,
−
4
i
^
+
2
j
^
−
k
^
-4\hat i+2\hat j-\hat k
−
4
i
^
+
2
j
^
−
k
^
respectively, then the projection of
O
P
⃗
\vec{OP}
O
P
on a vector perpendicular to the vectors
A
B
⃗
\vec{AB}
A
B
and
A
C
⃗
\vec{AC}
A
C
is:
A
10
3
\dfrac{10}{3}
3
10
B
8
3
\dfrac83
3
8
C
7
3
\dfrac73
3
7
D
3
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Official solution
A
B
⃗
=
(
4
,
3
,
1
)
\vec{AB}=(4,3,1)
A
B
=
(
4
,
3
,
1
)
,
A
C
⃗
=
(
−
2
,
1
,
2
)
\vec{AC}=(-2,1,2)
A
C
=
(
−
2
,
1
,
2
)
. \[\vec{AB}\times\vec{AC}=\hat i(5)-\hat j(8+2)+\hat k(4+6)=(5,-10,10).\] \[\text{Projection}=\frac{\vec{OP}\cdot(\vec{AB}\times\vec{AC})}{|\vec{AB}\times\vec{AC}|}=\frac{(-1)(5)+(-2)(-10)+(3)(10)}{\sqrt{25+100+100}}=\frac{45}{15}=3.\]
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