If-a-b-c-are-mutually-perpendicular-vectors-of-equal-magnitudes-show-that-the-vector-a-b-c-is-equally-inclined-to-a-b-and-c-

Question Number 15358 by Tinkutara last updated on 09/Jun/17

If \vec{a}, \vec{b}, \vec{c} are mutually perpendicular

vectors of equal magnitudes, show that

the vector \vec{a} + \vec{b} + \vec{c} is equally inclined

to \vec{a}, \vec{b} and \vec{c} .

Answered by prakash jain last updated on 09/Jun/17

a, b, c a r e m u t u a l p e r p e n d i c u l a r

(l i k e i, \bar{j}, k)

For any two vector x, y

x \cdot y =∣ x ∣∣ y ∣ \cos θ

where θ is angle between x, y

\cos θ = \frac{x \cdot y}{∣ x ∣∣∣ y ∣}

∣ a + b + c ∣= \sqrt{a^{2} + b^{2} + c^{2}}

(a + b + c) . a = a^{2}

(a + b + c) . b = b^{2}

(a + b + c) . c = c^{2}

f o r a n g l e θ_{a} b e t w e e n a + b + c a n d a

\cos θ_{a} = \frac{a^{2}}{a \sqrt{a^{2} + b^{2} + c^{2}}}

s i m i l a r l y θ_{b} a n d θ_{c}

\cos θ_{b} = \frac{b^{2}}{b \sqrt{a^{2} + b^{2} + c^{2}}}

\cos θ_{c} = \frac{c^{2}}{c \sqrt{a^{2} + b^{2} + c^{2}}}

s i n c e a = b = c

θ_{a} = θ_{b} = θ_{c}

Commented by Tinkutara last updated on 10/Jun/17

Thanks Sir!