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3-1.Vectors
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If three vectors along coordinate axis represent the adjacent sides of a cube of length $b$, then the unit vector along its diagonal passing through the origin will be
A$\frac{\hat{ i }+\hat{ j }+\hat{ k }}{\sqrt{2}}$
B$\frac{\hat{ i }+\hat{ j }+\hat{ k }}{\sqrt{36}}$
C$\hat{ i }+\hat{ j }+\hat{ k }$
D$\frac{\hat{ i }+\hat{ j }+\hat{ k }}{\sqrt{3}}$
Solution
(d)
Diagonal vector $A =b \hat{ i }+b \hat{ j }+b \hat{ k }$ or
$A =\sqrt{b^2+b^2+b^2}=\sqrt{3} b$
$\hat{ A }=\frac{ A }{A}=\frac{\hat{ i }+\hat{ j }+\hat{ k }}{\sqrt{3}}$
Diagonal vector $A =b \hat{ i }+b \hat{ j }+b \hat{ k }$ or
$A =\sqrt{b^2+b^2+b^2}=\sqrt{3} b$
$\hat{ A }=\frac{ A }{A}=\frac{\hat{ i }+\hat{ j }+\hat{ k }}{\sqrt{3}}$
Standard 11
Physics
