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(A) Given $a 	imes b = 2(a 	imes c)$,we can write $a 	imes (b - 2c) = 0$. This implies that the vector $(b - 2c)$ is parallel to $a$. Since $b - 2c

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given $a 	imes b = 2(a 	imes c)$,we can write $a 	imes (b - 2c) = 0$. This implies that the vector $(b - 2c)$ is parallel to $a$. Since $b - 2c = \lambda a$,we calculate the magnitude squared: $|b - 2c|^2 = |b|^2 + 4|c|^2 - 4(b \cdot c)$. Given $|b| = 4$,$|c| = 1$,and the angle $	heta$ between $b$ and $c$ is $\cos^{-1}\left(\frac{1}{4}ight)$,we have $b \cdot c = |b||c| \cos 	heta = 4 	imes 1 	imes \frac{1}{4} = 1$. Substituting these values: $|b - 2c|^2 = 4^2 + 4(1)^2 - 4(1) = 16 + 4 - 4 = 16$. Since $b - 2c = \lambda a$,we have $|\lambda a|^2 = 16$,which means $\lambda^2 |a|^2 = 16$. Given $|a| = 1$,we get $\lambda^2 = 16$,so $\lambda = \pm 4$. Considering the options provided,the correct value is $4$.

If $3$ vectors $a, b, c$ are such that $a \neq 0$ and $a \times b = 2(a \times c)$,$|a| = 1$,$|c| = 1$,$|b| = 4$,and the angle between $b$ and $c$ is $\cos^{-1}\left(\frac{1}{4}\right)$,and $b - 2c = \lambda a$,then $\lambda = $

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