Given a - -k-b - - -2 - k- c - -2-k- A unit vector perpendicular to both a - b and b - c is -

Let X be a vector perpendicular to both nbsp; a + b and b + c is: So, X = (a+ b) × (b+ c) X = (3 ĵ ) × ( 2 î + 4 ĵ) nbsp; = 6 k̂ Hence, unit vector is X̂ = X| ...

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Byju's Answer

Standard XII

Physics

Position Vector

Given a = î+ĵ...

Question

Given $\to a =^i +^j -^k, \to b = -^i + 2^j +^k, \to c = -^i + 2^j -^k .$ A unit vector perpendicular to both $\to a + \to b$ and $\to b + \to c$ is :

\frac{2^i +^j +^k}{\sqrt{6}}

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\frac{^i +^j +^k}{\sqrt{3}}

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Solution

The correct option is C $^k$
Let $\to X$ be a vector perpendicular to both $\to a + \to b$ and $\to b + \to c$ is:
So, $\to X = (\to a + \to b) \times (\to b + \to c)$
$\to X = (3^j) \times (- 2^i + 4^j) = 6^k$
Hence, unit vector is $^X = \frac{\to X}{| \to X |} = \frac{6^k}{6} =^k$

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Given →a=^i+^j−^k,→b=−^i+2^j+^k,→c=−^i+2^j−^k. A unit vector perpendicular to both →a+→b and →b+→c is :

The correct option is C ^kLet →X be a vector perpendicular to both →a+→b and →b+→c is: So, →X=(→a+→b)×(→b+→c) →X=(3^j)×(−2^i+4^j)=6^k Hence, unit vector is ^X=→X|→X|=6^k6=^k

Given $\to a =^i +^j -^k, \to b = -^i + 2^j +^k, \to c = -^i + 2^j -^k .$ A unit vector perpendicular to both $\to a + \to b$ and $\to b + \to c$ is :

The correct option is C $^k$
Let $\to X$ be a vector perpendicular to both $\to a + \to b$ and $\to b + \to c$ is:
So, $\to X = (\to a + \to b) \times (\to b + \to c)$
$\to X = (3^j) \times (- 2^i + 4^j) = 6^k$
Hence, unit vector is $^X = \frac{\to X}{| \to X |} = \frac{6^k}{6} =^k$