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Solution

Let $\vec{a} = (a_1, a_2, a_3)$ $\vec{a} \times \hat{i} = (0, a_3, -a_2)$ → magnitude$^2 = a_3^2 + a_2^2$ $\vec{a} \times \hat{j} = (-a_3, 0, a_1)$ → magnitude$^2 = a_3^2 + a_1^2$ $\vec{a} \times \hat{k} = (a_2, -a_1, 0)$ → magnitude$^2 = a_2^2 + a_1^2$ Sum = $2(a_1^2 + a_2^2 + a_3^2) = 2a^2$ $\boxed{\text{Answer: (D) }2a^2}$

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Solution

Vector perpendicular to both $\vec{a}$ and $\vec{b}$ is along $\vec{a} \times \vec{b}$. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1 \end{vmatrix} = (\hat{i})(1 - 2) - (\hat{j})(2 - 0) + (\hat{k})(2 - 0) = -\hat{i} - 2\hat{j} + 2\hat{k}$ Unit vector in this direction can be $\pm \dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$ → There are **two** such unit vectors. $\boxed{\text{Answer: (B) two}}$

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Solution

Using cross and dot product conditions, $\vec{B} = \hat{k}$.

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Solution

$\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{k} + \hat{j}$. The vector perpendicular to both is $\vec{a} \times \vec{b}$. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = \hat{i} - \hat{j} + \hat{k}$. Two unit vectors along $\pm(\hat{i} - \hat{j} + \hat{k})$ are possible.

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Solution

$\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$. Hence, angle between them is $180^{\circ}$.

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Solution

In a parallelogram, $A+C=B+D \Rightarrow D=A+C-B$. $D=(3,-1,2)+(-1,1,2)-(1,2,-4)=(1,-2,8)$.

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Solution

Dot product $= (5)(3) + (1)(-4) + (-3)(7)$ $= 15 - 4 - 21 = -10$ Wait — recheck: $15 - 4 - 21 = -10$ (not -15). Let’s verify carefully — if given answer key shows B (-15), check question once more: If second vector was $(3i - 4j + 7k)$ indeed, then $5×3 = 15$, $1×(-4) = -4$, $(-3)×7 = -21$, total = $-10$. So true answer is $-10$, though the paper marks B ($-15$). Maybe a misprint.

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Solution

Given $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$. Square both sides: $(\vec{a} + \vec{b})\cdot(\vec{a} + \vec{b}) = (\vec{a} - \vec{b})\cdot(\vec{a} - \vec{b})$ $\Rightarrow a^2 + b^2 + 2\vec{a}\cdot\vec{b} = a^2 + b^2 - 2\vec{a}\cdot\vec{b}$ $\Rightarrow \vec{a}\cdot\vec{b} = 0.$ Thus, $\vec{a}$ is perpendicular to $\vec{b}$.

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Solution

Equations of planes: $\pi_1: 2x + y + 2z = 8$ $\pi_2: 4x + 2y + 4z + 5 = 0$ Normalize the second plane by dividing by 2: $\pi_2: 2x + y + 2z + \dfrac{5}{2} = 0$ Distance between parallel planes $\dfrac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} = \dfrac{|8 - (-\tfrac{5}{2})|}{\sqrt{2^2 + 1^2 + 2^2}} = \dfrac{\tfrac{21}{2}}{\sqrt{9}} = \dfrac{3}{2}$ units.

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Solution

$\vec{a}\cdot\vec{b}$ is inner product.