If A, B, C, D are the points with position vectors ![]()
respectively, find the projection of
along
.
Given are points, A, B, C and D.
Let O be the origin.
We have,
Position vector of A![]()
![]()
Position vector of B![]()
![]()
Position vector of C![]()
![]()
Position vector of D![]()
![]()
Now, let us find out
and
.
Position vector of B-Position vector of A
![]()
![]()
![]()
![]()
And,
Position vector of D-Position vector of C
![]()
![]()
![]()
![]()
The projection of
along
is given by,



[∵ we know that,
and
.
So, ![]()
Also, we know that,
]
![]()
![]()
Multiply numerator and denominator by √21.

![]()
⇒ Projection = √21
Thus, projection of
along
is √21 units.
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