NIMCET Vector Previous Year Question Paper and Solution with PDF

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NIMCET
Previous Year Questions

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Previous Year Questions
NIMCET

NIMCET Previous Year Questions (PYQs)

NIMCET Vector PYQ

Qus : 1

$\vec{a}=4\hat{j}$ and

$\vec{b}=3\hat{j}+4\hat{k}$ , then the vector form of the component of

$\vec{a}$ alond

$\vec{b}$ is

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Solution

Qus : 2

$\vec{a}=\hat{i}-\hat{k}$ ,

$\vec{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}$ and

$\vec{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}$ , then

$\begin{bmatrix}{\vec{a}} & {\vec{b}} & {\vec{c}}\end{bmatrix}$ depends on

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Solution

Quick Solution

Given:

$\vec{a} = \hat{i} - \hat{k}, \quad \vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}, \quad \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$

Form the matrix:

$M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{bmatrix}$

Find the determinant:

$\det(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{vmatrix} = 1$

Since the determinant is constant and non-zero, the vectors are linearly independent.

$\boxed{\text{The matrix does not depend on } x \text{ or } y}$

Qus : 3

$\vec{a}$ and

$\vec{b}$ in space, given by

$\vec{a}=\frac{\hat{i}-2\hat{j}}{\sqrt{5}}$ and

$\vec{b}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt{14}}$ , then the value of

$(2\vec{a}+\vec{b}).[(\vec{a} \times \vec{b}) \times (\vec{a}-2\vec{b})]$ is

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Solution

Qus : 4

$\vec{a}, \vec{b}$ are unit vectors such that

$2\vec{a}+\vec{b} =3$ then which of the following statement is true?

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Solution

Quick Solution

Given: $\vec{a}, \vec{b}$ are unit vectors and

$2\vec{a} + \vec{b} = 3$

Take magnitude on both sides:

$|2\vec{a} + \vec{b}| = 3 \Rightarrow |2\vec{a} + \vec{b}|^2 = 9$

Use identity:

$|2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 + 4(\vec{a} \cdot \vec{b}) = 5 + 4(\vec{a} \cdot \vec{b})$

Set equal to 9:

$5 + 4(\vec{a} \cdot \vec{b}) = 9 \Rightarrow \vec{a} \cdot \vec{b} = 1 \Rightarrow \cos\theta = 1 \Rightarrow \theta = 0^\circ$

Qus : 5

$\theta={\cos }^{-1}\Bigg{(}\frac{3}{\sqrt[]{10}}\Bigg{)}$ is the angle between

$\vec{a}=\hat{i}-2x\hat{j}+2y\hat{k}$ &

$\vec{b}=x\hat{i}+\hat{j}+y\hat{k}$ then possible values of (x,y) that lie on the locus

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Solution

Qus : 6

If a vector having magnitude of 5 units, makes equal angle with each of the three mutually perpendicular axes,then the sum of the magnitude of the projections on each of the axis is

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Solution

Vector Projection Problem

Given: A vector of magnitude 5 makes equal angles with x, y, and z axes.
To Find: Sum of magnitudes of projections on each axis.

Let angle with each axis be $\alpha$ . Then, from direction cosine identity: $\cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \Rightarrow 3\cos^2\alpha = 1 \Rightarrow \cos\alpha = \frac{1}{\sqrt{3}}$

Projection on each axis: $5 \cdot \frac{1}{\sqrt{3}}$
Sum = $3 \cdot \frac{5}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \boxed{5\sqrt{3}}$

✅ Final Answer: $\boxed{5\sqrt{3}}$

Qus : 7

The value of non-zero scalars α and β such that for all vectors

and

such that

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Solution

Qus : 8

A force of 78 grams acts at the point (2,3,5). The direction ratios of the line of action being 2,2,1 . The magnitude of its moment about the line joining the origin to the point (12,3,4) is

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Solution

Qus : 9

The position vectors of the vertices

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Solution

Qus : 10

Not Available right now

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Solution

Qus : 11

Let

$\vec{a}, \vec{b}, \vec{c}$ be distinct non-negative numbers. If the vectors

$a\hat{i}+a\hat{j}+c\hat{k}$ ,

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

$c\hat{i}+c\hat{j}+b\hat{k}$ lie in a plane, then c is

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Solution

$\vec{a}=a\hat{i}+a\hat{j}+c\hat{k}\, ,\, \vec{b}=\hat{i}+\hat{k}\, \&\, \vec{c}=c\hat{i}+c\hat{j}+b\hat{k}$ are coplanar.

$\Rightarrow\begin{vmatrix}{a} & {a} & {c} \\ {1} & {0} & {1} \\ {c} & {c} & {b}\end{vmatrix}=0$

$\Rightarrow-ac-ab+ac+{c}^2=0$

$\Rightarrow{c}^2=ab$

Qus : 12

The value of m for which volume of the parallelepiped is 4 cubic units whose three edges are represented by a = mi + j + k, b = i – j + k, c = i + 2j –k is

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Solution

Given: Volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is 4 cubic units.

Vectors:

$\vec{a} = m\hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} - \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$

Step 1: Volume = $|\vec{a} \cdot (\vec{b} \times \vec{c})|$

First compute $\vec{b} \times \vec{c}$ :

$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(2) - (-1)(1)) \\ = \hat{i}(1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 + 1) = -\hat{i} + 2\hat{j} + 3\hat{k}$

Step 2: Compute dot product with $\vec{a}$ :

$\vec{a} \cdot (\vec{b} \times \vec{c}) = (m)(-1) + (1)(2) + (1)(3) = -m + 2 + 3 = -m + 5$

Step 3: Volume = $| -m + 5 | = 4$

So, $|-m + 5| = 4 \Rightarrow -m + 5 = \pm 4$

Case 1: $-m + 5 = 4 \Rightarrow m = 1$
Case 2: $-m + 5 = -4 \Rightarrow m = 9$

✅ Final Answer: $\boxed{m = 1 \text{ or } 9}$

Qus : 13

The number of distinct real values of

$\lambda$ for which the vectors

${\lambda}^2\hat{i}+\hat{j}+\hat{k},\, \hat{i}+{\lambda}^2\hat{j}+j$ and

$\hat{i}+\hat{j}+{\lambda}^2\hat{k}$ are coplanar is

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Solution

Given: Vectors:

$\vec{a} = \lambda^2 \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + \lambda^2 \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + \hat{j} + \lambda^2 \hat{k}$

Condition: Vectors are coplanar ⟹ Scalar triple product = 0

$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$

Step 1: Use determinant:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} \lambda^2 & 1 & 1 \\ 1 & \lambda^2 & 1 \\ 1 & 1 & \lambda^2 \end{vmatrix}$

Step 2: Expand the determinant:

$= \lambda^2(\lambda^2 \cdot \lambda^2 - 1 \cdot 1) - 1(1 \cdot \lambda^2 - 1 \cdot 1) + 1(1 \cdot 1 - \lambda^2 \cdot 1) \\ = \lambda^2(\lambda^4 - 1) - (\lambda^2 - 1) + (1 - \lambda^2)$

Simplify:

$= \lambda^6 - \lambda^2 - \lambda^2 + 1 + 1 - \lambda^2 = \lambda^6 - 3\lambda^2 + 2$

Step 3: Set scalar triple product to 0:

$\lambda^6 - 3\lambda^2 + 2 = 0$

Step 4: Let $x = \lambda^2$ , then:

$x^3 - 3x + 2 = 0$

Factor:

$x^3 - 3x + 2 = (x - 1)^2(x + 2)$

So, $\lambda^2 = 1$ (double root), or $\lambda^2 = -2$ (discard as it's not real)

Thus, real values of $\lambda$ are: $\lambda = \pm1$

✅ Final Answer: $\boxed{2}$ distinct real values

Qus : 14

If the volume of the parallelepiped whose adjacent edges are

$\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$ ,

$\vec{b}=\hat{i}+\alpha \hat{j}+2\hat{k}$ and

$\vec{c}=\hat{i}+2\hat{j}+\alpha \hat{k}$ is 15, then

$\alpha$ is equal to

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Solution

Qus : 15

If F|= 40N (Newtons), |D| = 3m, and

$\theta={60^{\circ}}$ , then the work done by F acting

from P to Q is

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Solution

Formula for work done:

$W = |F| \cdot |D| \cdot \cos\theta$

Given:

$|F| = 40 \, \text{N}$
$|D| = 3 \, \text{m}$
$\theta = 60^\circ$

Step 1: Plug in the values:

$W = 40 \cdot 3 \cdot \cos(60^\circ)$

Step 2: Use $\cos(60^\circ) = \frac{1}{2}$

$W = 40 \cdot 3 \cdot \frac{1}{2} = 60 \, \text{J}$

✅ Final Answer: $\boxed{60 \, \text{J}}$

Qus : 16

Let

$\vec{a}=2\hat{i}+2\hat{j}+\hat{k}$ and

$\vec{b}$ be another vector such that

$\vec{a}.\vec{b}=14$ and

$\vec{a} \times \vec{b}=3\hat{i}+\hat{j}-8\hat{k}$ the vector

$\vec{b}$ =

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Solution

Qus : 17

A man starts at the origin O and walks a distance of 3 units in the north- east direction and then walks a distance of 4 units in the north-west direction to reach the point P. then

$\vec{OP}$ is equal to

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Solution

A man starts at the origin $O$ , walks 3 units in the north-east direction, then 4 units in the north-west direction to reach point $P$ . Find the displacement vector $\vec{OP}$ .

? Solution:

North-East (45°): $\vec{A} = 3 \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \right)$
North-West (135°): $\vec{B} = 4 \cdot \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( -\frac{4}{\sqrt{2}}, \frac{4}{\sqrt{2}} \right)$
Total Displacement: $\vec{OP} = \vec{A} + \vec{B} = \left( \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right)$

✅ Final Answer:

$\boxed{ \vec{OP} = \left( \frac{-1}{\sqrt{2}},\ \frac{7}{\sqrt{2}} \right) }$

Qus : 18

$\vec{a}=\lambda \hat{i}+\hat{j}-2\hat{k}$ ,

$\vec{b}=\hat{i}+\lambda \hat{j}-2\hat{k}$ and

$\vec{c}=\hat{i}+\hat{j}+\hat{k}$ and

$\begin{bmatrix}{\vec{a}} & {\vec{b}} & {\vec{c}} \end{bmatrix}=7$ , then the values of the

$\lambda$ are

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Solution

Qus : 19

How much work does it take to slide a crate for a distance of 25m along a loading dock by pulling on it with a 180 N force where the dock is at an angle of 45° from the horizontal?

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Solution

Work Done Problem:

A crate is pulled 25 m along a dock with a force of 180 N at an angle of 45°.

✅ Formula Used:

$\text{Work} = F \cdot d \cdot \cos(\theta)$

✅ Substituting Values:

$W = 180 \times 25 \times \cos(45^\circ) = 180 \times 25 \times 0.70710678118 = 3181.98052\, \text{J}$

✅ Final Answer (to 5 decimal places):

$\boxed{3.181\times 10^3 \, \text{Joules}}$

Qus : 20

$(\vec{a} \times \vec{b}) \times \vec{c}= \vec{a} \times (\vec{b} \times \vec{c})$ , then

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Solution

Qus : 21

The length of the projection of

$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ on

$\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$ , is equal to:

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Solution

Qus : 22

Let

$\vec{a}=2\widehat{i}\, +\widehat{j}\, +2\widehat{k}$ ,

$\vec{b}=\widehat{i}-\widehat{j}+2\widehat{k}$ and

$\vec{c}=\widehat{i}+\widehat{j}-2\widehat{k}$ are are three vectors. Then, a vector in the plane of

$\vec{a}$ and

$\vec{c}$ whose projection on

$\vec{b}$ is of magnitude

$\frac{1}{\sqrt{6}}$ is

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Solution

Qus : 23

If the position vector of A and B relative to O be

$\widehat{i}\, -4\widehat{j}+3\widehat{k}$ and

$-\widehat{i}\, +2\widehat{j}-\widehat{k}$ respectively, then the median through O of ΔABC is:

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Solution

Qus : 24

The area of the triangle formed by the vertices whose position vectors are

$3\widehat{i}+\widehat{j}$ ,

$5\widehat{i}+2\widehat{j}+\widehat{k}$ ,

$\widehat{i}-2\widehat{j}+3\widehat{k}$ is

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Solution

Qus : 25

$\vec{a}, \vec{b}$ and

$\vec{c}$ are three vectors such that

$\vec{a} \times \vec{b}=\vec{c}$ ,

$\vec{a}.\vec{c} = 2$ and

$\vec{b}.\vec{c} = 1$ . If

$|\vec{b}| = 1$ , then the value of

$|\vec{a}|$ is

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Solution

Qus : 26

Let

$\vec{a}$ ,

$\vec{b}$ and

$\vec{c}$ be unit vectors such that the angle between them is

${\cos }^{-1}\Bigg{\{}\frac{1}{4}\Bigg{\}}$ . If

$\vec{b}=2\vec{c}+\lambda \vec{a}$ , where

$\lambda$ > 0 and

$\vec{b}=4$ , then

$\lambda$ is equal to

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Solution

Qus : 27

$\vec{a}$ and

$\vec{b}$ are twp vectors such that |

$\vec{a}$ |=3, |

$\vec{b}$ |=4 and |

$\vec{a}+\vec{b}$ |=1, then the value of

$|\vec{a}-\vec{b}|$ is

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Solution

Qus : 28

$\vec{a}=\hat{i}+\hat{j}+\hat{k}$ ,

$\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$ and

$\vec{c}=\hat{i}-2\hat{j}+\hat{k}$ , then a vector of magnitude

$\sqrt{22}$ which is parallel to

$2\vec{a}-\vec{b}+3\vec{c}$ is

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Solution

Qus : 29

If the vectors

$a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k},\hat{i}+\hat{j}+c\hat{k}$ ,

$(a,b,c\ne1)$ are coplanar, then

$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$

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Solution

Qus : 30

The direction cosines of the vector a = (- 2i + j – 5k) are

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Solution

Qus : 31

Let

$\vec{a}=\hat{i}+\hat{j}$ and

$\vec{b}=2\hat{i}-\hat{k}$ , the point of intersection of the lines

$\vec{r}\times\vec{a}=\vec{b}\times\vec{a}$ and

$\vec{r}\times\vec{b}=\vec{a}\times\vec{b}$ is

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Solution

Qus : 32

$\vec{a}$ ,

$\vec{b}$ and

$\vec{c}$ are vectors such that

$\vec{a}$ +

$\vec{b}$ +

$\vec{c}$ = 0 and |

$\vec{a}$ | =7,

$\vec{b}$ =5, |

$\vec{c}$ | = 3, then the angle between the vectors

$\vec{b}$ and

$\vec{c}$

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Solution

Qus : 33

and

, (a ≠ b ≠ c ≠ 1) are co-planar, then the value of

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Solution

Qus : 34

Let a, b and c be three vectors having magnitudes 1, 1 and 2 respectively. If a x (a x c) - b = 0, then the acute angle between a and c is

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Solution

Qus : 35

Let

and

be three vector such that |

| = 2, |

| = 3, |

| = 5 and

= 0. The value of

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Solution

Qus : 36

Constant forces

$\vec{P}= 2\hat{i} - 5\hat{j} + 6\hat{k}$ and

$\vec{Q}= -\hat{i} + 2\hat{j}- \hat{k}$ act on a particle. The work done when the particle is displaced from A whose position vector is

$4\hat{i} - 3\hat{j} - 2\hat{k}$ , to B whose position vector is

$6\hat{i} + \hat{j} - 3k\hat{k}$ , is:

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Solution

Qus : 37

and

are unit vectors, then

does not exceeds

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Solution

Qus : 38

= (i + 2j - 3k) and

=(3i -j + 2k), then the angle between (

) and (

)

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Solution

Qus : 39

Let

$\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$ ,

$\vec{b}=\hat{i}+2\hat{j}-\hat{k}$ and

$\vec{c}=3\hat{i}+\hat{j}+\lambda \hat{k}$ be the co-terminal edges

of a parallelopiped whose volume is 5 units. Then the value of

$\lambda$ is

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Solution

Qus : 40

The vector

lies in the plane of the vector

and

and bisects the angle between

and

. Then which of the following gives possible values of

and

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Solution

Qus : 41

For the vectors

$\vec{a}=-4\hat{i}+2\hat{j}, \vec{b}=2\hat{i}+\hat{j}$ and

$\vec{c}=2\hat{i}+3\hat{j}$ , if

$\vec{c}=m\vec{a}+n\vec{b}$ then the value of m + n is

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Solution

Qus : 42

A bird is flying in a straight line with velocity vector 10i+6j+k, measured in km/hr. If the starting point is (1,2,3), how much time does it to take to reach a point in space that is 13m high from the ground?

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Solution

Qus : 43

Angle between

$\vec{a}$ and

$\vec{b}$ is

$120{^{\circ}}$ . If

$|\vec{b}|=2|\vec{a}|$ and the vectors ,

$\vec{a}+x\vec{b}$ ,

$\vec{a}-\vec{b}$ are at right angle, then

$x=$

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Solution

Qus : 44

Let

$\vec{a}$ and

$\vec{b}$ be two vectors, which of the following vectors are not perpendicular to each other?

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4
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Solution

Qus : 45

$\vec{e_1}=(1,1,1)$ and

$\vec{e_2}=(1,1,-1)$ and

$\vec{a}$ and

$\vec{b}$ and two vectors such that

$\vec{e_2}=\vec{a}+2\vec{b}$ , then angle between

$\vec{a}$ and

$\vec{b}$

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Solution

Qus : 46

$\vec{A}=4\hat{i}+3\hat{j}+\hat{k}$ and

$\vec{B}=2\hat{i}-\hat{j}+2\hat{k}$ , then the unit vector

$\hat{N}$ perpendicular to the vectors

$\vec{A}$ and

$\vec{B}$ ,such that

$\vec{A}, \vec{B}$ , and

$\hat{N}$ form a right handed system, is:

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Solution

Qus : 47

The sum of two vectors

$\vec{a}$ and

$\vec{b}$ is a vector

$\vec{c}$ such that

$|\vec{a}|=|\vec{b}|=|\vec{c}|=2$ . Then, the magnitude of

$\vec{a}-\vec{b}$ is equal to:

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Solution

Qus : 48

$\vec{a}=\hat{i}-\hat{k},\, \vec{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}$ and

$\vec{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}$ , then

$[\vec{a} , \vec{b}, \vec{c}]$ depends on

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Solution

Qus : 49

Let

$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and

$\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. A vector

$\vec{v}$ in the plane of

$\vec{a}$ and

$\vec{b}$ whose projection on

$\frac{\vec{c}}{|\vec{c}|}$ is

$\frac{1}{\sqrt{3}}$ , is

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4
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Solution

Qus : 50

$\vec{a}, \vec{b}$ and

$\vec{c}$ are the position vectors of the vertices A, B, C of a triangle ABC, then the area of the triangle ABC is

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Solution

Qus : 51

If a vector

$\vec{a}$ makes an equal angle with the coordinate axes and has magnitude 3, then the angle between

$\vec{a}$ and each of the three coordinate axes is

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Solution

Qus : 52

A cube is made up of 125 one cm. square cubes placed on a table. How many squares are visible only on three sides?

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Solution

Qus : 53

$\vec{AC}=2\hat{i}+\hat{j}+\hat{k}$ and

$\vec{BD}=-\hat{i}+3\hat{j}+2\hat{k}$ then the area of the quadrilateral ABCD is

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Solution

Qus : 54

are three non-coplanar vectors, then

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Solution

Qus : 55

Two forces F₁ and F₂ are used to pull a car, which met an accident. The angle between the two forces is θ . Find the values of θ for which the resultant force is equal to

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Solution

Qus : 56

are four vectors such that

is collinear with

and

is collinear with

then

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Solution

Qus : 57

Forces of magnitude 5, 3, 1 units act in the directions 6i + 2j + 3k, 3i - 2j + 6k, 2i - 3j - 6k respectively on a particle which is displaced from the point (2, −1, −3) to (5, −1, 1). The total work done by the force is

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Solution

Qus : 58

The position vectors of points A and B are

and

. Then the position vector of point p dividing AB in the ratio m : n is

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Solution

Qus : 59

If a, b, c are three non-zero vectors with no two of which are collinear, a + 2b is collinear with c and b + 3c is collinear with a , then | a + 2b + 6c | will be equal to

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Solution

Qus : 60

Vertices of the vectors i - 2j + 2k , 2i + j - k and 3i - j + 2k form a triangle. This triangle is

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4
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Solution

Qus : 61

If the volume of a parallelepiped whose adjacent edges are

a = 2i + 3j + 4k,

b = i + αj + 2k

c = i + 2k + αk

is 15, then α =

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4
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Solution

Qus : 62

$\overrightarrow{{a}}$ and

$\overrightarrow{{b}}$ are vectors in space, given by

$\overrightarrow{{a}}=\frac{\hat{i}-2\hat{j}}{\sqrt[]{5}}$ and

$\overrightarrow{{b}}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt[]{14}}$ , then the value of

$(2\vec{a} + \vec{b}).[(\vec{a} × \vec{b}) × (\vec{a} – 2\vec{b})]$ is

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4
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Solution

Qus : 63

Let

$\vec{A} = 2\hat{i} + \hat{j} – 2\hat{k}$ and

$\vec{B} = \hat{i} + \hat{j}$ , If

$\vec{C}$ is a vector such that

$|\vec{C} – \vec{A}| = 3$ and the angle between A × B and C is

${30^{\circ}}$ , then

$|(\vec{A} × \vec{B}) × \vec{C}|$ = 3 then the value of

$\vec{A}.\vec{C}$ is equal to

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4
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Solution

Qus : 64

$\vec{a}$ and

$\vec{b}$ are vectors such that

$|\vec{a}|=13$ ,

$|\vec{b}|=5$ and

$\vec{a} . \vec{b} =60$ then the value of

$|\vec{a} \times \vec{b}|$ is

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4
Go to Discussion

Solution

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NIMCET Previous Year Questions (PYQs)

NIMCET Vector PYQ

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Vector Projection Problem

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Given: Volume of a parallelepiped formed by vectors →a,→b,→c\vec{a}, \vec{b}, \vec{c} is 4 cubic units.

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Given: Vectors:

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A man starts at the origin O O , walks 3 units in the north-east direction, then 4 units in the north-west direction to reach point P P . Find the displacement vector →OP \vec{OP} .

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Given: Volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is 4 cubic units.

A man starts at the origin $O$ , walks 3 units in the north-east direction, then 4 units in the north-west direction to reach point $P$ . Find the displacement vector $\vec{OP}$ .