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Joint Entrance Examination
Mathematics
Differential Equations
Question 223
Question 223 of 490
Easy
MCQ
*(Vector geometry --- not a differential equation; included as it appears in the source under this heading.)* In
△
A
B
C
\triangle ABC
△
A
B
C
, if
B
C
=
8
BC=8
B
C
=
8
,
C
A
=
7
CA=7
C
A
=
7
,
A
B
=
10
AB=10
A
B
=
10
, the projection of
A
B
⃗
\vec{AB}
A
B
on
A
C
⃗
\vec{AC}
A
C
is:
A
25
4
\dfrac{25}{4}
4
25
B
85
14
\dfrac{85}{14}
14
85
C
127
20
\dfrac{127}{20}
20
127
D
115
16
\dfrac{115}{16}
16
115
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Official solution
Projection
=
A
B
cos
θ
=
10
⋅
A
B
2
+
A
C
2
−
B
C
2
2
⋅
A
B
⋅
A
C
=
100
+
49
−
64
14
=
85
14
=AB\cos\theta = 10\cdot\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC} = \dfrac{100+49-64}{14} = \dfrac{85}{14}
=
A
B
cos
θ
=
10
⋅
2
⋅
A
B
⋅
A
C
A
B
2
+
A
C
2
−
B
C
2
=
14
100
+
49
−
64
=
14
85
.
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