Vector Algebra Question 141
Question: Three forces of magnitudes 1, 2, 3 dynes meet in a point and act along diagonals of three adjacent faces of a cube. The resultant force is
[MNR 1987]
Options:
A) 114 dyne
B) 6 dyne
C) 5 dyne
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Three vectors meeting at a point are $ \mathbf{i}+\mathbf{j},,\mathbf{j}+\mathbf{k},,\mathbf{k}+\mathbf{i} $ . Forces of 1, 2, 3 dynes are acting along these directions respectively, therefore resultant force
= $ \frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}}+\frac{2(\mathbf{j}+\mathbf{k})}{\sqrt{2}}+\frac{3(\mathbf{k}+\mathbf{i})}{\sqrt{2}}=\frac{4\mathbf{i}+3\mathbf{j}+5\mathbf{k}}{\sqrt{2}} $ ,
$ \therefore $ Magnitute = $ \frac{5\sqrt{2}}{\sqrt{2}}= $ 5 dyne.
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