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Let $α, β, γ$ be distinct real numbers. The points with position vectors $α \hat{i} + β \hat{j} + γ \hat{k}, β \hat{i} + γ \hat{j} + α \hat{k}, γ \hat{i} + α \hat{j} + β \hat{k}$

are collinear

from an equilateral triangle

form an isosceles triangle

from a right angled triangle

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Two vectors $\vec{A}$ and $\vec{B}$ lie in a plane. Another vector $\vec{C}$ lies outside this plane. The resultant $\vec{A} + \vec{B} + \vec{C}$ of these three vectors

can be zero

cannot be zero

lies in the plane of $\vec{A} & \vec{B}$

lies in the plane of $\vec{A} & \vec{A} + \vec{B}$

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Explanation

If A and B lie in a plane, and C lies outside this plane, then A + B + C cannot be zero. This is because the sum of any two vectors lying in a plane must also lie in that plane, and adding a non-zero vector from outside the plane cannot make the resultant zero.

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A set of vectors taken in a given order gives a closed polygon. Then the resultant of these vectors is a

scalar quantity

pseudovector

unit vector

null vector

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Explanation

When a set of vectors are added in a closed polygon, their resultant is a null vector. This is because the vectors form a closed loop, and their sum cancels out to zero.

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The vector sum of two P and Q is minimum when the angle $θ$ between their positive directions, is

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

$π$

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Explanation

(D)

$R = \sqrt{{(P)}^{2} + {(Q)}^{2} + 2 P Q \cos θ} W h e n θ = π, \cos θ = - 1 t h e n R i s m i n i m u m$

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The vector sum of two vectors $\vec{A}$ and $\vec{B}$ is maximum, then the angle $θ$ between two vectors is -

$0 °$

$30 °$

$45 °$

$60 °$

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Explanation

$R^{2} = P^{2} + Q^{2} + 2 P Q \cos θ W h e n θ = 0, \cos = 1 S o R i s m a x i m u m w h e n θ = 0$

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Given: $\vec{C} = \vec{A} + \vec{B}$ . Also, the magnitude of $\vec{A}, \vec{B}$ and $\vec{C}$ are 12, 5 and 13 units respectively. The angle between $\vec{A}$ and $\vec{B}$ is

$0 °$

$\frac{π}{4}$

$\frac{π}{2}$

$π$

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Explanation

$A s \vec{C} = \vec{A} + \vec{B}$

Let angle between A and B is $θ$

$\Rightarrow C^{2} = A^{2} + B^{2} + 2 A B C o s θ \Rightarrow {(13)}^{2} = {(12)}^{2} + {(5)}^{2} + 2 \times 12 + 5 + \cos θ \Rightarrow \cos θ = 0 \Rightarrow θ = 90 °$

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If $\vec{P} + \vec{Q} = \vec{P} - \vec{Q}$ and $θ$ is the angle between $\vec{P}$ and $\vec{Q}$ , then

$θ = 0 °$

$θ = 90 °$

$P = 0$

$Q = 0$

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Explanation

$\vec{P} + \vec{Q} = \vec{P} - \vec{Q}$

$2 \vec{Q} = 0$

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What is the torque of a force $F = (2 \hat{i} - 3 \hat{j} + 4 \hat{k})$ Newton acting at a point $r = (3 \hat{i} + 2 \hat{j} + 3 \hat{k})$ metre about the origin?

$6 \hat{i} - 6 \hat{j} + 12 \hat{k}$

$17 \hat{i} - 6 \hat{j} - 13 \hat{k}$

$- 6 \hat{i} + 6 \hat{j} - 12 \hat{k}$

$- 17 \hat{i} + 6 \hat{j} - 13 \hat{k}$

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Explanation

(B)

$\vec{τ} = \vec{r} \times \vec{F} = (3 \hat{i} + 2 \hat{j} + 3 \hat{k}) \times (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & - 3 & 4 \end{matrix}| = (2 \times 4 - (- 3) \times 3 \hat{i}) + (3 \times 4 - 3 \times 2) (- \hat{j}) + (3 \times - 3 - 2 \times 2) \hat{k} = 17 \hat{i} - 6 \hat{j} - 13 \hat{k}$

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Three non zero vectors $\vec{A}, \vec{B} & \vec{C}$ satisfy the relation $\vec{A} \cdot \vec{B} = 0 & \vec{A} \cdot \vec{C} = 0$ . Then $\vec{A}$ can be parallel to:

$\vec{B}$

$\vec{C}$

$\vec{B} \cdot \vec{C}$

$\vec{B} \times \vec{C}$

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Explanation

$(4) As \vec{A} . \vec{B} = 0 and \vec{A} . \vec{C} = 0, so \vec{A} is ⊥ ar to both \vec{B} and \vec{C} . Also, \vec{B} \times \vec{C} is also ⊥ ar to both \vec{B} and \vec{C} . So, \vec{A} can be parallel to \vec{B} \times \vec{C} .$

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Given: $\vec{a} + \vec{b} + \vec{c} = 0$ . Out of the three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ two are equal in magnitude. The magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. The angles between the vectors are:

90 $°$ , 135 $°$ , 135 $°$

30 $°$ , 60 $°$ , 90 $°$

45 $°$ , 45 $°$ , 90 $°$

45 $°$ , 60 $°$ , 90 $°$

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