3-1.Vectors
medium

सदिशों $\mathop A\limits^ \to = 3\hat i + 4\hat j + 5\hat k$ तथा $\mathop B\limits^ \to = 3\hat i + 4\hat j - 5\hat k$ के बीच का कोण....... $^o$ है

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

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