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Three forces starts acting simultaneously on a particle moving with velocity $\vec{v} .$ These forces are represented in magnitude and direction by the three sides of a triangle $A B C$ (as shown). The particle will now move with velocity

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Important Questions on Vector Algebra

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

3 \hat{i} + 2 \hat{j} - 5 \hat{k} = x (2 \hat{i} - \hat{j} + \hat{k}) + y (\hat{i} + 3 \hat{j} - 2 \hat{k}) + z (- 2 \hat{i} + \hat{j} - 3 \hat{k})

,

then

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

If the sum of two unit vectors is again a unit vector, then magnitude of their difference is

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

\hat{i} + 4 \hat{j} + 3 \hat{k}, \hat{i} + 2 \hat{j} + 3 \hat{k}, 3 \hat{i} + 2 \hat{j} + \hat{k}

are position vectors of

A, B, C

respectively and if

D, E

are mid points of sides

B C

and

A C

, then

\vec{D E}

is equal to

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

If two forces of $3$ units and $4$ units are acting at an angle $90 °$ , then its resultant forces will be

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

If the vectors

\vec{AB} = 3 \hat{i} + 4 \hat{k}

and

\vec{AC} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}

are the sides of a triangle

A B C,

then the length of the median through

A

is:

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

\vec{a} = 2 \hat{i} + 5 \hat{j}

and

\vec{b} = 2 \hat{i} - \hat{j}

, then the unit vector along

\vec{a} + \vec{b}

will be

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

The diagonals of a parallelogram are the vectors

3 \hat{i} + 6 \hat{j} - 2 \hat{k}

and

- \hat{i} - 2 \hat{j} - 8 \hat{k}

then, the length of the shorter side of parallelogram is

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

In quadrilateral

A B C D, \vec{A B} = \vec{a}, \vec{B C} = \vec{b}, \vec{D A} = \vec{a} - \vec{b}, M

is the midpoint of

B C

and

X

is a point on

D M

such that

\vec{D X} = \frac{4}{5} \vec{D M}

. Then the points

A, X

and

C

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

In a

∆ A B C,

D, E, F

are the mid-points of the sides

B C

C A

and

A B

respectively, the vector

\vec{A D}

is equal to

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

Let

u

and

v

be two vectors. Then,

|u - v| = | |u| - |v| |

if and only if

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

The two adjacent sides of parallelogram are

2 \hat{i} - 4 \hat{j} + 5 \hat{k}

and

\hat{i} - 2 \hat{j} - 3 \hat{k}

. The unit vector parallel to its diagonal which starts at the common vertex of the above two sides, is

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

\vec{a}

and

\vec{b}

are unit vectors, then the greatest value of

\sqrt{3} | \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

\vec{A} = 3 \hat{i} - 2 \hat{j} + \hat{k}, \vec{B} = \hat{i} - 3 \hat{j} + 5 \hat{k}

and

\vec{C} = 2 \hat{i} + \hat{j} - 4 \hat{k}

form a right angled triangle then out of the following which one is satisfied?

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

Let

\vec{v}

be a vector in the plane such that

| \vec{v} - \vec{i} | = | \vec{v} - 2 \vec{i} | = | \vec{v} - \vec{j} | .

Then

| \vec{v} |

lies in the interval.

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

\vec{a} = 2 \hat{ı} + 3 \hat{ȷ} - 4 \hat{k}

and

\vec{b} = \hat{ı} + 3 \hat{ȷ} + 2 \hat{k}

, then a unit vector in the direction of

\vec{a} + \vec{b}

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

Prove by vector method that: $\cos (A - B) = \cos A \cos B + \sin A \sin B$ .

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

The resultant of two forces $P$ and $Q$ is $R$ . If $Q$ is doubled, then the new resultant will be perpendicular to $P$ . Prove that $Q = R$ .

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

The two vectors

\hat{i} + \hat{j} + \hat{k}

and

\hat{i} + 3 \hat{j} + 5 \hat{k}

represent the two sides

\vec{A B}

and

\vec{A C}

respectively of a

Δ A B C

. The length of the median through

A

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

In a triangle $P Q R$ , let $\vec{a} = \vec{Q R}, \vec{b} = \vec{R P}$ and $\vec{c} = \vec{P Q}$ .

If $|\vec{a}| = 3, |\vec{b}| = 4$ and $\frac{\vec{a} . (\vec{c} - \vec{b})}{\vec{c} . (\vec{a} - \vec{b})} = \frac{|\vec{a}|}{|\vec{a}| + |\vec{b}|}$ , then the value of ${|\vec{a} \times \vec{b}|}^{2}$ is _______

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Addition of Vectors

Let

\vec{P R} = 3 \hat{i} + \hat{j} - 2 \hat{k} & \vec{S Q} = \hat{i} - 3 \hat{j} - 4 \hat{k}

determine diagonals of a parallelogram

P Q R S

and

\vec{P T} = \hat{i} + 2 \hat{j} + 3 \hat{k}

be another vector. Then the volume (in cubic unit) of the parallelepiped determined by the vectors

\vec{P T}, \vec{P Q} & \vec{P S}

Three forces starts acting simultaneously on a particle moving with velocity v→. These forces are represented in magnitude and direction by the three sides of a triangle ABC (as shown). The particle will now move with velocity

Important Questions on Vector Algebra

Learn and Prepare for any exam you want !

Three forces starts acting simultaneously on a particle moving with velocity $\vec{v} .$ These forces are represented in magnitude and direction by the three sides of a triangle $A B C$ (as shown). The particle will now move with velocity