a → = 2 i ^ + 3 j ^ + 4 k ^ & b → = 4 i ^ + 2 j ^ + 3 k ^ . Prove Cauchy-Schwarz inequality for vectors.

Given that, a → = 2 i ^ + 3 j ^ + 4 k ^ & b → = 4 i ^ + 2 j ^ + 3 k ^ . Cauchy-Schwarz inequality for vectors states that, a → · b → ≤ a → b → . a → · b → = 2 × 4 + 3 × 2 + 4 × 3 = 26 | ( a → · b → ) | = 26 = 26 | a → | = 2 2 + 3 2 + 4 2 = 29 | b → | = 4 2 + 2 2 + 3 2 = 29 Verifying, a → · b → ≤ a → b → ⇒ 26 ≤ 29 × 29 ⇒ 26 ≤ 29 Hence, proved Cauchy-Schwarz inequality for vectors.

a → = 2 i ^ + 3 j ^ + 4 k ^ & b → = 4 i ^ + 2 j ^ + 3 k ^ . Prove Cauchy-Schwarz inequality for vectors.

Given that, a → = 2 i ^ + 3 j ^ + 4 k ^ & b → = 4 i ^ + 2 j ^ + 3 k ^ . Cauchy-Schwarz inequality for vectors states that, a → · b → ≤ a → b → . a → · b → = 2 × 4 + 3 × 2 + 4 × 3 = 26 | ( a → · b → ) | = 26 = 26 | a → | = 2 2 + 3 2 + 4 2 = 29 | b → | = 4 2 + 2 2 + 3 2 = 29 Verifying, a → · b → ≤ a → b → ⇒ 26 ≤ 29 × 29 ⇒ 26 ≤ 29 Hence, proved Cauchy-Schwarz inequality for vectors.

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

Earn 100

$\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} & \vec{b} = 4 \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Prove Cauchy-Schwarz inequality for vectors.

Important Questions on Vector Algebra

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 :

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

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?

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}

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

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

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-

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\vec{a}, \vec{b}

and

\vec{c}

, be three unit vectors such that

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

. If

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

and

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

, then the order pair,

(λ, \vec{d})

, 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

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

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

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

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

\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}

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

\vec{P} and \vec{Q}

are two non-zero vectors inclined to each other at an angle

θ .

\hat{p} and \hat{q}

are unit vectors along

\vec{P} a n d \vec{Q}

respectively. The component of

\vec{Q}

in the direction of

\vec{P}

will be

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

If the angle between

\vec{a}

and

\vec{b}

\frac{2 π}{3}

and the projection of

\vec{a}

in the direction of

\vec{b}

- 2,

then

| \vec{a} | =

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

If $| \vec{a} | = 10, | \vec{b} | = 2$ and $\vec{a} \cdot \vec{b} = 12$ , then the value of $| \vec{a} \times \vec{b} |$ is

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

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} = 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})

a→=2i^+3j^+4k^ & b→=4i^+2j^+3k^. Prove Cauchy-Schwarz inequality for vectors.

Important Questions on Vector Algebra

Learn and Prepare for any exam you want !

$\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} & \vec{b} = 4 \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Prove Cauchy-Schwarz inequality for vectors.