If x- is a vector such that - p-q-r- -x- p-q- -r- then x- is

The correct option is C indeterminate, as [p⃗q⃗r⃗] Let p=i̅ q=j̅ r=k̅ Hence p̅×q̅ =k Now #160; [(p̅×q̅).k̅].x̅=(p̅×q̅)×r̅ ...

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Byju's Answer

Standard XII

Physics

Vectors and Its Types

If x⃗ is a ...

Question

If $\to x$ is a vector such that $[\to p \to q \to r] \to x = (\to p \times \to q) \times \to r,$ then $\to x$ is

c (^i - 2^j +^k)

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a unit vector

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indeterminate, as

[\to p \to q \to r]

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- \frac{1}{2} (^i - 2^j +^k)

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Solution

The correct option is C indeterminate, as $[\to p \to q \to r]$
Let $p = ¯ i$
$q = ¯ j$
$r = ¯ k$
Hence
$¯ p \times ¯ q$
$= k$
Now
$[(¯ p \times ¯ q) . ¯ k] . ¯ x = (¯ p \times ¯ q) \times ¯ r$
$[¯ k . ¯ k] ¯ x = (¯ k \times ¯ k)$
Thus vector (x) turns out to be 0.
Hence the value of vector x is determined by the chosen values of vectors, p, q, and r.
Unless the value of p,q, and r are not fixed, we cannot quantify x.
Hence the correct answer is 'indeterminate'.

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If →x is a vector such that [→p→q→r]→x=(→p×→q)×→r, then →x is

If $\to x$ is a vector such that $[\to p \to q \to r] \to x = (\to p \times \to q) \times \to r,$ then $\to x$ is