Vector Algebra Question 388
Question: If $ l,\mathbf{a}+m,\mathbf{b}+n,\mathbf{c}=\mathbf{0}, $ where $ l,,m,n $ are scalars and a, b, c are mutually perpendicular vectors, then
Options:
A) $ l=m=n=1 $
B) $ l+m+n=1 $
C) $ l=m=n=0 $
D) $ l\ne 0,m\ne 0,n\ne 0 $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ l\mathbf{a}+m\mathbf{b}+n\mathbf{c}=0 $ Squaring both sides, we get $ a^{2}l^{2}+m^{2}b^{2}+n^{2}c^{2}+2l,m,\mathbf{a},.,\mathbf{b}+2l,n,\mathbf{a},.,\mathbf{c}+2m,n,\mathbf{b},.,\mathbf{c}=0 $ But $ \mathbf{a},,\mathbf{b},,\mathbf{c} $ are mutually perpendicular So, $ \mathbf{a},.,\mathbf{b}, $ $ \mathbf{b},.,\mathbf{c} $ and $ \mathbf{c},.,\mathbf{a} $ are equal to zero. Therefore, $ a^{2}l^{2}+m^{2}b^{2}+n^{2}c^{2}=0i.e.,l,m,n $ are equal to zero because $ a^{2},b^{2} $ and $ c^{2} $ cannot be equal to zero.
BETA
