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3-1.Vectors
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The angle between the two vectors $\vec A = 3\hat i + 4\hat j + 5\hat k$ and $\vec B = 3\hat i + 4\hat j - 5\hat k$ will be....... $^o$
A
$90$
B
$0$
C
$60$
D
$45$
(AIPMT-1994)
Solution
(a) $\cos \theta = \frac{{\overrightarrow A .\overrightarrow B }}{{|A||B|}} = \frac{{(3\hat i + 4\hat j + 5\hat k)\,(3\hat i + 4\hat j – 5\hat k)}}{{\sqrt {9 + 16 + 25} \sqrt {9 + 16 + 25} }}$
$ = \frac{{9 + 16 – 25}}{{50}} = 0$
$⇒$ $\cos \theta = 0$,
$\therefore $ $\theta = 90^\circ $
Standard 11
Physics
