Let u-v- and w-2-3k- If n- is a unit vector such that u-n-0 and v-n-0-then -w-n-

$\to V = 2^i +^j -^k$ and $\to W =^i + 3^k$ . If $\to U$ is a unit vector, then the maximum value of the scalar triple product $[\to U \to V \to W]$ is

Q. If

\to u =^j + 4^k

\to v =^i - 3^k

, and

\to w = cos θ^i + sin θ^j

are vectors in

3

-dimensional space, then the maximum possible value of

| \to u \times \to v \cdot \to w |

is?

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Let →u=^i+^j,→v=^i−^j and →w=^i+2^j+3^k. If ^n is a unit vector such that →u.^n=0 and →v.^n=0,then |→w.^n|=

The correct option is C 3Let the vector be xi+yj+zkThen as u.n=0,x+y=0v.n=0,x−y=0Thus, x=y=0As n is a unit vector the z-component has to be unity only.So |n.w|=3

Let $\to u =^i +^j, \to v =^i -^j$ and $\to w =^i + 2^j + 3^k$ . If $^n$ is a unit vector such that $\to u .^n = 0$ and $\to v .^n = 0$ ,then $| \to w .^n | =$

The correct option is C $3$
Let the vector be $x i + y j + z k$
Then as $u . n = 0, x + y = 0$
$v . n = 0, x - y = 0$
Thus, $x = y = 0$
As $n$ is a unit vector the $z$ -component has to be unity only.
So $| n . w | = 3$