In a triangle A B C , right angle at vertex A , if the position vectors of A , B and C are respectively 3 i ^ + j ^ - k ^ , - i ^ + 3 j ^ + p k ^ and 5 i ^ + q j ^ - 4 k ^ , then the point p , q lies on a line:

Making an acute angle with the positive direction of x - a x i s

Making an obtuse angle with the positive direction of x - a x i s

EASY

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If $\vec{A} \cdot \vec{B} = \vec{A} \times \vec{B}$ then angle between $\vec{A}$ and $\vec{B}$ is:

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Important Questions on Vector Algebra

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

\vec{a}

and

\vec{b} = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}

are collinear and

\vec{a} . \vec{b} = 27,

then

\vec{a}

is equal to

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

|\vec{a} - \vec{b}| = |\vec{a}| = |\vec{b}| = 1

,

then the angle between

\vec{a}

and

\vec{b}

is equal to

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

If the vectors

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

and

\vec{c} = λ \hat{i} + 9 \hat{j} + μ \hat{k}

are mutually orthogonal, then

λ + μ

is equal to

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\vec{v_{1}}, \vec{v_{2}}, \vec{v_{3}}, \vec{v_{4}}

be unit vectors in the

x y

- plane, one each in the interior of the four quadrants. Which of the following statements is NOT necessarily true?

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

If vectors

\vec{A} = \cos ω t \hat{i} + s i n ω t \hat{j}

and

\vec{B} = c o s \frac{ω t}{2} \hat{i} + \sin \frac{ω t}{2} \hat{j}

are functions of time, then the value of

t

at which they are orthogonal to each other is:

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\vec{a} = (6 \hat{i} - 3 \hat{j} - 6 \hat{k})

and

\vec{d} = (\hat{i} + \hat{j} + \hat{k})

. Suppose that

\vec{a} = \vec{b} + \vec{c}

where

\vec{b}

is parallel to

\vec{d}

and

\vec{c}

is perpendicular to

\vec{d}

. Then

\vec{c}

is-

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

|\vec{a}| = 3, |\vec{b}| = 1, |\vec{c}| = 4

and

\vec{a} + \vec{b} + \vec{c} = 0

,

then the value of

\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}

is equal to

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

\vec{a}, \vec{b}, \vec{c}

are vectors such that

\vec{a} + \vec{b} + \vec{c} = 0

and

|\vec{a}| = 7, |\vec{b}| = 5, |\vec{c}| = 3

then angle between vector

\vec{b}

and

\vec{c}

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

In a parallelogram

A B C D, |\vec{A B}| = a, |\vec{A D}| = b & |\vec{A C}| = c

\vec{D B} \cdot \vec{A B}

has the value:

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\vec{A} = (\hat{i} + \hat{j}) a n d \vec{B} = (2 \hat{i} - \hat{j})

. The magnitude of a coplanar vector

\vec{C}

such that

\vec{A} . \vec{C} = \vec{B} . \vec{C} = \vec{A} . \vec{B}

is given by:

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

|\vec{a}| = 2, |\vec{b}| = 3

and

|2 \vec{a} - \vec{b}| = 5,

then

|2 \vec{a} + \vec{b}|

equals :

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\vec{a} = 2 \hat{i} + λ_{1} \hat{j} + 3 \hat{k}, \vec{b} = 4 \hat{i} + (3 - λ_{2}) \hat{j} + 6 \hat{k}

and

\vec{c} = 3 \hat{i} + 6 \hat{j} + (λ_{3} - 1) \hat{k}

be three vectors such that

\vec{b} = 2 \vec{a}

and

\vec{a}

is perpendicular to

\vec{c} .

Then a possible value of

(λ_{1}, λ_{2}, λ_{3})

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

\vec{x} = 3 \hat{i} - 6 \hat{j} - \hat{k}, \vec{y} = \hat{i} + 4 \hat{j} - 3 \hat{k}

and

\vec{z} = 3 \hat{i} - 4 \hat{j} - 12 \hat{k}

, then the magnitude of the projection of

\vec{x} \times \vec{y}

\vec{z}

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

The vectors

3 \vec{a} - 5 \vec{b}

and

2 \vec{a} + \vec{b}

are mutually perpendicular and the vectors

\vec{a} + 4 \vec{b}

and

- \vec{a} + \vec{b}

are also mutually perpendicular. Then the angle between the vectors

\vec{a}

and

\vec{b}

, is

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

In a triangle

A B C

, right angle at vertex

A

, if the position vectors of

A, B

and

C

are respectively

3 \hat{i} + \hat{j} - \hat{k}, - \hat{i} + 3 \hat{j} + p \hat{k}

and

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

, then the point

(p, q)

lies on a line:

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

The projection of the line segment joining the points

(1, - 1, 3)

and

(2, - 4, 11)

on the line joining the points

(- 1, 2, 3)

and

(3, - 2,10)

is _______

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

θ

is the angle between two vectors

\vec{a}

and

\vec{b},

then

\vec{a} \cdot \vec{b} \geq 0

only when

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Given, $\vec{a} = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} .$ Let $\vec{c}$ be a vector such that $|\vec{c} - \vec{a}| = 3, |(\vec{a} \times \vec{b}) \times \vec{c}| = 3$ and the angle between $\vec{c}$ and $\vec{a} \times \vec{b}$ be $30 °$ . Then $\vec{a} \cdot \vec{c}$ is equal to:

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\vec{a} = \hat{i} + \hat{j} + \sqrt{2} \hat{k},

\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + \sqrt{2} \hat{k}

and

\vec{c} = 5 \hat{i} + \hat{j} + \sqrt{2} \hat{k}

be three vectors such that the projection vector of

\vec{b}

\vec{a}

|\vec{a}|

. If

\vec{a} + \vec{b}

is perpendicular to

\vec{c}

, then

|\vec{b}|

is equal to:

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Two vectors

\vec{A}

and

\vec{B}

have equal magnitudes. The magnitude of

(\vec{A} + \vec{B})

n

times the magnitude of

(\vec{A} - \vec{B})

. The angle between

\vec{A}

and

\vec{B}

is:

If A → · B → = A → × B → then angle between A → and B → is:

Important Questions on Vector Algebra

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If $\vec{A} \cdot \vec{B} = \vec{A} \times \vec{B}$ then angle between $\vec{A}$ and $\vec{B}$ is: