If a vector a of magnitude 50 is collinear wi | vedclass

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(B) Let $a = xi + yj + zk$. Given $|a| = 50$ and $b = 6i - 8j - \frac{15}{2}k$.Since $a$ and $b$ are collinear, $a = \lambda b$ for some scalar $\lamb

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$-24i + 32j + 30k$

Vedclass · Answer

(B) Let $a = xi + yj + zk$. Given $|a| = 50$ and $b = 6i - 8j - \frac{15}{2}k$.Since $a$ and $b$ are collinear, $a = \lambda b$ for some scalar $\lambda$.Thus, $a = \lambda(6i - 8j - \frac{15}{2}k) = 6\lambda i - 8\lambda j - \frac{15}{2}\lambda k$.Given $|a| = 50$, we have $|a|^2 = 2500$.$|a|^2 = (6\lambda)^2 + (-8\lambda)^2 + (-\frac{15}{2}\lambda)^2 = 36\lambda^2 + 64\lambda^2 + \frac{225}{4}\lambda^2 = 100\lambda^2 + \frac{225}{4}\lambda^2 = \frac{400 + 225}{4}\lambda^2 = \frac{625}{4}\lambda^2$.Setting $\frac{625}{4}\lambda^2 = 2500$, we get $\lambda^2 = \frac{2500 	imes 4}{625} = 4 	imes 4 = 16$, so $\lambda = \pm 4$.Since $a$ makes an acute angle with the positive $z$-axis, the $z$-component of $a$ must be positive.The $z$-component is $-\frac{15}{2}\lambda$. For this to be positive, $\lambda$ must be negative, so $\lambda = -4$.Substituting $\lambda = -4$ into $a = \lambda b$, we get $a = -4(6i - 8j - \frac{15}{2}k) = -24i + 32j + 30k$.

If a vector $a$ of magnitude $50$ is collinear with the vector $b = 6i - 8j - \frac{15}{2}k$, and makes an acute angle with the positive direction of the $z$-axis, then the vector $a$ is equal to:

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