Given that →A→A = ˆiˆi - 2ˆjˆj + 3ˆkˆk and →B→B = 3ˆiˆi - 4ˆkˆk, find the projection of A on B. (Ans: -5/9 )
Given:
→A→A = ˆiˆi - 2ˆjˆj + 3ˆkˆk →B→B = 3ˆiˆi - 4ˆkˆk
→A→A = ˆiˆi - 2ˆjˆj + 3ˆkˆk
→B→B = 3ˆiˆi - 4ˆkˆk
To Find:
The projection of vector →A→A on →B→B = A cos 𝞡 = ?
Solution:
We know that the projection of →A→A is on →B→B is A cos 𝜭, So, the scalar (dot) product is
→B→B . →A→A = B (A cos 𝜭) or
A cos 𝜭 = →B.→AB→B.→AB
A cos 𝜭 = (3ˆi-4ˆk).(ˆi-2ˆj+3ˆk)√Bx2+By2+Bz2(3ˆi−4ˆk).(ˆi−2ˆj+3ˆk)√Bx2+By2+Bz2
A cos 𝜭 = (3x1)ˆi.ˆi+(3x-2)ˆi.ˆj+(3x3)ˆj.ˆk+(-4x1)ˆk.ˆi+(-4x-2)ˆk.ˆj+(-4x3)ˆk.ˆk√32+(0)2+(-4)2(3x1)ˆi.ˆi+(3x−2)ˆi.ˆj+(3x3)ˆj.ˆk+(−4x1)ˆk.ˆi+(−4x−2)ˆk.ˆj+(−4x3)ˆk.ˆk√32+(0)2+(−4)2
[∴ ˆiˆi . ˆiˆi = ˆjˆj . ˆjˆj = ˆkˆk . ˆkˆk =1andˆiˆi . ˆjˆj = ˆjˆj . ˆiˆi = ˆjˆj . ˆkˆk = ˆkˆk . ˆjˆj = ˆkˆk . ˆiˆi = ˆiˆi . ˆkˆk= 0 ]
by putting the above values and calculating
A cos 𝜭 = 3-12√9+163−12√9+16
A cos 𝜭 = -9√25−9√25
A cos 𝜭 = -95−95 ------ Ans
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Numerical Problem 2.10 ⇑ |
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