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(A) Let $\vec{r}_1 = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{r}_2 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.Given $a^2 + b^2 + c^2 = 1$,we have $|\vec{r}_1|

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(A) Let $\vec{r}_1 = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{r}_2 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.Given $a^2 + b^2 + c^2 = 1$,we have $|\vec{r}_1| = 1$.The cross product is given by $\vec{r}_1 	imes \vec{r}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a & b & c \ 2 & 3 & 4 \end{vmatrix} = \hat{i}(4b - 3c) - \hat{j}(4a - 2c) + \hat{k}(3a - 2b)$.The square of the magnitude is $|\vec{r}_1 	imes \vec{r}_2|^2 = (4b - 3c)^2 + (4a - 2c)^2 + (3a - 2b)^2$.Using the property $|\vec{r}_1 	imes \vec{r}_2| = |\vec{r}_1| |\vec{r}_2| \sin 	heta$,where $	heta$ is the angle between the vectors.Since $|\vec{r}_1| = 1$ and $|\vec{r}_2| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$,we have $|\vec{r}_1 	imes \vec{r}_2| = \sqrt{29} \sin 	heta$.Therefore,$|\vec{r}_1 	imes \vec{r}_2|^2 = 29 \sin^2 	heta$.The maximum value of $\sin^2 	heta$ is $1$,so the maximum value of the expression is $29 	imes 1 = 29$.

Let $a$,$b$,and $c$ be real numbers such that $a^2 + b^2 + c^2 = 1$. Then the maximum value of $(4b - 3c)^2 + (4a - 2c)^2 + (3a - 2b)^2$ is:

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