If a-b-c are three mutually perpendicular vectors equally inclined to a-b-c at angle - then find the value of 36cos22-

Finding the value of 36cos^22θ:Since the vectors are mutually perpendicular,a.b=b.c=c.a =0(a+b+c).( a+b+c) =| a+b+c|^20ex0ex⇒|a|^2+|b|^2+|c|^2=| a+b+c|^2… (1)0e ...

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Standard XII

Mathematics

Angle between Two Vectors

If a,b,c are ...

Question

If $\vec{a,} \vec{b}, \vec{c}$ are three mutually perpendicular vectors equally inclined to $\vec{a} + \vec{b} + \vec{c}$ at angle $θ$ then find the value of $36 \cos^{2} 2 θ$ .

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Solution

Finding the value of $36 \cos^{2} 2 θ$ :
Since the vectors are mutually perpendicular,
$\vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0$
$(\vec{a} + \vec{b} + \vec{c}) . (\vec{a} + \vec{b} + \vec{c}) = {|\vec{a} + \vec{b} + \vec{c}|}^{2} \Rightarrow {|\vec{a}|}^{2} + | {\vec{b |}}^{2} + {|\vec{c}|}^{2} = {|\vec{a} + \vec{b} + \vec{c}|}^{2} \dots (1) \Rightarrow \vec{a} . (\vec{a} + \vec{b} + \vec{c}) = | \vec{a} | | \vec{a} + \vec{b} + \vec{c} | \cos θ \Rightarrow {|\vec{a}|}^{2} = | \vec{a} | | \vec{a} + \vec{b} + \vec{c} | \cos θ \Rightarrow | \vec{a} | = | \vec{a} + \vec{b} + \vec{c} | \cos θ$
In the same manner,
$|\vec{b}| = |\vec{a} + \vec{b} + \vec{c}| \cos θ | \vec{c} | = |\vec{a} + \vec{b} + \vec{c}| \cos θ \Rightarrow |\vec{a}| = |\vec{b}| = |\vec{c}|$
From equation $(1)$ , we get
$3 | \vec{a} |^{2} = | \vec{a} + \vec{b} + \vec{c} |^{2} \Rightarrow 3 {|\vec{a} + \vec{b} + \vec{c}|}^{2} \cos^{2} θ = | \vec{a} + \vec{b} + \vec{c} |^{2} \Rightarrow \cos^{2} θ = \frac{1}{3} \Rightarrow 1 + \cos 2 θ = 2 \cos^{2} θ \Rightarrow 36 \cos^{2} 2 θ = 36 {(2 \times 1 / 3 - 1)}^{2} \Rightarrow = \frac{36}{9} \Rightarrow = 4$
Hence, the required answer is 4.

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If a,→b→,c→ are three mutually perpendicular vectors equally inclined to a→+b→+c→ at angle θ then find the value of 36cos22θ.

If $\vec{a,} \vec{b}, \vec{c}$ are three mutually perpendicular vectors equally inclined to $\vec{a} + \vec{b} + \vec{c}$ at angle $θ$ then find the value of $36 \cos^{2} 2 θ$ .