If p-2-10-2 - q-3-1-2 and r-2-1-3 - then -p-q-r-

The correct option is C 0 p⃗=2#160;i 10#160;j+2#160;k q⃗=3#160;i+#160;j+2#160;k r⃗=2#160;i+#160;j+3#160;k Now p⃗× (q⃗×r⃗) =q⃗(p⃗. ...

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Byju's Answer

Standard XII

Physics

Vector Addition

If p⃗=2,-10...

Question

If $\to p = (2, - 10, 2)$ , $\to q = (3, 1, 2)$ and $\to r = (2, 1, 3)$ , then $| \to p \times (\to q \times \to r) | =$

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Solution

The correct option is C $0$
$\to p = 2^i - 10^j + 2^k$
$\to q = 3^i +^j + 2^k$
$\to r = 2^i +^j + 3^k$
Now $\to p \times (\to q \times \to r)$
$= \to q (\to p . \to r) - \to r (\to p . \to q)$
Now $\to p . \to r = 4 - 10 + 6$
$= 0$
And $\to p . \to q = 6 - 10 + 4$
$= 0$ .
Therefore
$\to p \times (\to q \times \to r)$
$= \to q (\to p . \to r) - \to r (\to p . \to q)$
$= \to q (0) - \to r (0)$
$= 0$

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If →p=(2,−10,2) , →q=(3,1,2) and →r=(2,1,3) , then |→p×(→q×→r)|=

The correct option is C 0→p=2^i−10^j+2^k→q=3^i+^j+2^k→r=2^i+^j+3^kNow →p×(→q×→r)=→q(→p.→r)−→r(→p.→q)Now →p.→r=4−10+6=0And →p.→q=6−10+4=0.Therefore→p×(→q×→r)=→q(→p.→r)−→r(→p.→q)=→q(0)−→r(0)=0

If $\to p = (2, - 10, 2)$ , $\to q = (3, 1, 2)$ and $\to r = (2, 1, 3)$ , then $| \to p \times (\to q \times \to r) | =$