Let the vectors a-b-c be such that - a-2- -b-4 and -c-4- If the projection of b on a is equal to the projection of c on a and b is perpendicular to c- then the value of -a-b-c- is

⇒b·a2 = c·a2 ⇒b·a =c·a ⋯ (i) And b.c= 0 (as perpendiculars) ⋯ (ii) ⇒ |a+b c nbsp;|=√(a^2+b^2+c^2+2a.b 2b.c 2a.c) =√(4+16+16) = 6 ...

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Standard XII

Physics

Summary

Let the vecto...

Question

Let the vectors $\to a, \to b, \to c$ be such that $| \to a | = 2,$ $| \to b | = 4$ and | $\to c | = 4.$ If the projection of $\to b$ on $\to a$ is equal to the projection of $\to c$ on $\to a$ and $\to b$ is perpendicular to $\to c$ , then the value of $| \to a + \to b - \to c$ | is

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Solution

$\Rightarrow \frac{\to b \cdot \to a}{2} = \frac{\to c \cdot \to a}{2}$
$\Rightarrow \to b \cdot \to a = \to c \cdot \to a \dots (i)$

And $\to b . \to c = 0 (as perpendiculars) \dots (i i)$

$\Rightarrow | \to a + \to b - \to c | = \sqrt{a^{2} + b^{2} + c^{2} + 2 \to a . \to b - 2 \to b . \to c - 2 \to a . \to c}$
$= \sqrt{4 + 16 + 16}$
= 6

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Let the vectors →a,→b,→c be such that |→a|=2, |→b|=4 and |→c|=4. If the projection of →b on →a is equal to the projection of →c on →a and →b is perpendicular to →c, then the value of |→a+→b−→c| is

⇒→b⋅→a2=→c⋅→a2 ⇒→b⋅→a=→c⋅→a ⋯(i) And →b.→c=0 (as perpendiculars) ⋯(ii) ⇒|→a+→b−→c|=√a2+b2+c2+2→a.→b−2→b.→c−2→a.→c =√4+16+16 = 6

Let the vectors $\to a, \to b, \to c$ be such that $| \to a | = 2,$ $| \to b | = 4$ and | $\to c | = 4.$ If the projection of $\to b$ on $\to a$ is equal to the projection of $\to c$ on $\to a$ and $\to b$ is perpendicular to $\to c$ , then the value of $| \to a + \to b - \to c$ | is