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Physics MCQs for NEET — Practice Questions with Answers

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A vector that is perpendicular to both the vectors $\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is

$- \hat{i} + \hat{k}$

$- \hat{i} - 2 \hat{j} + \hat{k}$

$\hat{i} - 2 \hat{j} + \hat{k}$

$\hat{i} + \hat{k}$

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Explanation

(A)

$\hat{i} + \hat{k} = - \hat{i} + 0 \hat{j} + \hat{k} \vec{a} . (- \hat{i} + \hat{k}) = - 1 + 0 + 1 = 0 \vec{b} . (- \hat{i} + \hat{k}) = - 1 + 0 + 1 = 0$

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If $θ$ is the angle between vectors $\vec{a}$ and $\vec{b}$ , and $| \vec{a} \times \vec{b} | = \vec{a} . \vec{b}$ , then $θ$ is equal to

$0 °$

$180 °$

$135 °$

$45 °$

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Explanation

$G i v e n, | \vec{a} \times \vec{b} | = \vec{a} . \vec{b} \Rightarrow a b \sin θ = a b \cos θ \Rightarrow \tan θ = 1 \Rightarrow θ = 45 °$

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The vector $\vec{b}$ , which is collinear with the vector $\vec{a}$ =(2, 1, -1) and satisfies the condition $\vec{a}$ . $\vec{b}$ =3, is

(1, 1/2, -1/2)

(2/3, 1/3, -1/3)

(1/2, 1/4, -1/4)

(1, 1, 0)

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Explanation

$Going through the options; Only (A) satifies the condition \vec{a} . \vec{b} = 3 .$

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If $3 \hat{i} + 2 \hat{j} + 8 \hat{k}$ and $2 \hat{i} + x \hat{j} + \hat{k}$ are at right angle then x=

-7

-4

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Explanation

If two vectors are at right angle, then their dot product = 0

$\Rightarrow (3 \hat{i} + 2 \hat{j} + 8 \hat{k}) . (2 \hat{i} + x \hat{j} + \hat{k}) = 0 \Rightarrow 6 + 2 x + 8 = 0 \Rightarrow x = - 7$

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If a, b, c are three non-zero vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$ the value of $\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}$ is

Less than zero

equal to zero

greater than zero

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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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