If $a, b, c$ are three vectors such that $| a | = 3, | b | = 4, | c | = 5$ and $a, b, c$ are perpendicular to $b + c, c + a, a + b$ respectively, then $| a + b + c | =$

6 \sqrt{2}

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4 \sqrt{2}

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3 \sqrt{2}

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5 \sqrt{2}

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Solution

The correct option is D $5 \sqrt{2}$
$∵ a ⊥ (b + c)$
$∴ a . (b + c) = 0 \Rightarrow a . b + a . c = 0$ ...(1)
Similarly $b ⊥ (c + a) \Rightarrow b . c + b . a = 0$ ...(2)
and $c ⊥ (a + b) = 0 \Rightarrow c . a + c . b = 0$ ...(3)
Adding (1),(2),(3), we get
$2 (a . b + b . c + c . a) = 0$
Now, ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a . b + b . c + c . a)$
$= {| a |}^{2} + {| b |}^{2} + {| c |}^{2} + 0 = 3^{2} + 4^{2} + c^{2} = 50$
Hence $| a + b + c | = 5 \sqrt{2}$

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If a,b,c are three vectors such that |a|=3,|b|=4,|c|=5 and a,b,c are perpendicular to b+c,c+a,a+b respectively, then |a+b+c|=

The correct option is D 5√2∵a⊥(b+c)∴a.(b+c)=0⇒a.b+a.c=0 ...(1)Similarly b⊥(c+a)⇒b.c+b.a=0 ...(2)and c⊥(a+b)=0⇒c.a+c.b=0 ...(3)Adding (1),(2),(3), we get2(a.b+b.c+c.a)=0Now, (a+b+c)2=a2+b2+c2+2(a.b+b.c+c.a)=|a|2+|b|2+|c|2+0=32+42+c2=50Hence |a+b+c|=5√2

If $a, b, c$ are three vectors such that $| a | = 3, | b | = 4, | c | = 5$ and $a, b, c$ are perpendicular to $b + c, c + a, a + b$ respectively, then $| a + b + c | =$