If the vector vec{a} = (x, y, z) makes an obt | vedclass

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(B) Given $\vec{a} = (x, y, z)$. Since $\vec{a}$ makes an obtuse angle with the $y$-axis,the $y$-component must be negative,i.e.,$y < 0$.It is given t

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$(2, -2, -2)$

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(B) Given $\vec{a} = (x, y, z)$. Since $\vec{a}$ makes an obtuse angle with the $y$-axis,the $y$-component must be negative,i.e.,$y It is given that $\vec{a}$ makes equal angles with $\vec{b}$ and $\vec{c}$,so $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{\vec{a} \cdot \vec{c}}{|\vec{a}| |\vec{c}|}$.Note that $|\vec{b}| = \sqrt{y^2 + 4z^2 + 9x^2}$ and $|\vec{c}| = \sqrt{4z^2 + 9x^2 + y^2}$,so $|\vec{b}| = |\vec{c}|$.Thus,$\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$.$xy - 2yz + 3zx = 2zx + 3xy - yz$.$2xy + yz - zx = 0$ ... $(i)$.Also,$\vec{a} \cdot \vec{d} = 0$,so $x - y + 2z = 0$,which gives $z = \frac{y - x}{2}$ ... $(ii)$.Substituting $(ii)$ into $(i)$:$2xy + y(\frac{y - x}{2}) - x(\frac{y - x}{2}) = 0$.$4xy + y^2 - xy - xy + x^2 = 0$.$x^2 + 2xy + y^2 = 0 \Rightarrow (x + y)^2 = 0 \Rightarrow y = -x$.From $(ii)$,$z = \frac{-x - x}{2} = -x$.So,$\vec{a} = (x, -x, -x)$.Given $|\vec{a}| = 2\sqrt{3}$,we have $\sqrt{x^2 + (-x)^2 + (-x)^2} = 2\sqrt{3} \Rightarrow |x|\sqrt{3} = 2\sqrt{3} \Rightarrow |x| = 2$.Since $y Thus,$x = 2, y = -2, z = -2$.Therefore,$\vec{a} = (2, -2, -2)$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

If the vector $\vec{a} = (x, y, z)$ makes an obtuse angle with the $y$-axis and makes equal angles with the vectors $\vec{b} = (y, -2z, 3x)$ and $\vec{c} = (2z, 3x, -y)$,and if $|\vec{a}| = 2\sqrt{3}$ and $\vec{a}$ is perpendicular to $\vec{d} = (1, -1, 2)$,find the vector $\vec{a}$.

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