Let α → = 3 i ^ + j ^ and β → = 2 i ^ - j ^ + 3 k ^ . If β → = β 1 → - β 2 → , where β 1 → is parallel to α → and β 2 → is perpendicular to α → , then β 1 → × β 2 → is equal to:

1 2 ( - 3 i ^ + 9 j ^ + 5 k ^ )

HARD

Earn 100

If $\vec{x}$ and $\vec{y}$ be two non-collinear vectors and a triangle $A B C$ with side lengths $a, b, c$ satisfy the vector equation
$(20 a - 15 b) \vec{x} + (15 b - 12 c) \vec{y} + (12 c - 20 a) (\vec{x} \times \vec{y}) = \vec{0}$ ,
then $∆ A B C$ is

50% studentsanswered this correctly

Important Questions on Vector Algebra

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

The magnitude of the projection of the vector

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

on the vector perpendicular to the plane containing the vectors

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

and

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

is:

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\vec{a}, \vec{b}

and

\vec{c}

be three vectors such that

|\vec{a}| = \sqrt{3}, |\vec{b}| = 5, \vec{b} \cdot \vec{c} = 10

and the angle between

\vec{b}

and

\vec{c}

\frac{π}{3} .

\vec{a}

is perpendicular to the vector

\vec{b} \times \vec{c},

then

|\vec{a} \times (\vec{b} \times \vec{c})|

is equal to ____________.

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\vec{α} = 3 \hat{i} + \hat{j}

and

\vec{β} = 2 \hat{i} - \hat{j} + 3 \hat{k} .

\vec{β} = \vec{β_{1}} - \vec{β_{2}},

where

\vec{β_{1}}

is parallel to

\vec{α}

and

\vec{β_{2}}

is perpendicular to

\vec{α},

then

\vec{β_{1}} \times \vec{β_{2}}

is equal to:

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

A unit vector perpendicular to the plane containing the vectors

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

and

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

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

|\vec{a}| = 13, |\vec{b}| = 5

and

\vec{a} \cdot \vec{b} = 30

,

then

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

is equal to

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

| \vec{a} | = 16, | \vec{b} | = 4

then,

\sqrt{| \vec{a} \times \vec{b} |^{2} + | \vec{a} \cdot \vec{b} |^{2}} =

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

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

and

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

for some real

x

. Then

|\vec{a} \times \vec{b}| = r

is possible if

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

The area of the parallelogram, whose diagonals are

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

and

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

is equal to

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

The area (in sq. units) of the parallelogram whose diagonals are along the vectors

8 \hat{i} - 6 \hat{j}

and

3 \hat{i} + 4 \hat{j} - 12 \hat{k},

is:

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

If the vector

\vec{b} = 3 \hat{j} + 4 \hat{k}

is written as the sum of a vector

\vec{b_{1}}

, parallel to

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

and a vector

{\vec{b}}_{2},

perpendicular to

\vec{a},

then

{\vec{b}}_{1} \times {\vec{b}}_{2}

is equal to :

MEDIUM

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let the position vectors of points

A

and

B

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

and

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

respectively. A point

P

divides the line segment

A B

internally in the ratio

λ : 1 (λ > 0)

. If

O

is the origin and

\vec{OB} \cdot \vec{OP} - 3 {|\vec{OA} \times \vec{OP}|}^{2} = 6

, then

λ

is equal to

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:

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

{(\vec{a} \times \vec{b})}^{2} + {(\vec{a} \cdot \vec{b})}^{2} = 676

and

|\vec{b}| = 2

, then

|\vec{a}|

is equal to

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

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

and

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

be two vectors. If a vector perpendicular to both the vectors

\vec{a} + \vec{b}

and

\vec{a} - \vec{b}

has the magnitude

12

then one such vector is:

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\hat{a}

and

\hat{b}

are two non-collinear unit vectors. If

\vec{u} = \hat{a} - (\hat{a} \cdot \hat{b}) \hat{b}

and

\vec{v} = \hat{a} \times \hat{b}

, then

| \vec{v} |

is equal to

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

A vector

\vec{a}

of length

2

units is making an angle

60 °

with each of the

X

-axis and

Y

-axis. If another vector

\vec{b}

of length

\sqrt{2}

units is making an angle

45 °

with each of the

Y

-axis and

Z

-axis, then

\vec{a} \times \vec{b} =

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

\vec{a} + l \vec{b} + l^{2} \vec{c} = 0

and

\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 3 (\vec{b} \times \vec{c}),

then the minimum value of such

l

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

\vec{a} = \hat{\dot{i}} + \hat{\dot{j}} + \hat{k}, \vec{b} = \hat{\dot{i}} + 3 \hat{\dot{j}} + 5 \hat{k}

and

\vec{c} = 7 \hat{\dot{i}} + 9 \hat{\dot{j}} + 11 \hat{k}

then the area of parallelogram having diagonals

\vec{a} + \vec{b}

and

\vec{b} + \vec{c}

EASY

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Let

\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + 3 \hat{j} + 5 \hat{k}

and

\vec{c} = 7 \hat{i} + 9 \hat{j} + 11 \hat{k}

.

Then, the area of the parallelogram with diagonals

\vec{a} + \vec{b}

and

\vec{b} + \vec{c}

HARD

Mathematics>Coordinate Geometry>Vector Algebra>Product of Vectors

Consider the cube in the first octant with sides

O P, O Q

and

O R

of length

1

, along the

x

-axis,

y

-axis and

z

-axis, respectively, where

O (0, 0, 0)

is the origin. Let

S (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})

be the centre of the cube and

T

be the vertex of the cube opposite to the origin

O

such that

S

lies on the diagonal

O T

. If

\vec{p} = \vec{S P}, \vec{q} = \vec{S Q}, \vec{r} = \vec{S R}

and

\vec{t} = \vec{S T}

, then the value of

|(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})|

is ______.

If x→ and y→ be two non-collinear vectors and a triangle ABC with side lengths a,b,c satisfy the vector equation 20a-15bx→+15b-12cy→+12c-20ax→×y→=0→, then ∆ABC is

Important Questions on Vector Algebra

Learn and Prepare for any exam you want !

If $\vec{x}$ and $\vec{y}$ be two non-collinear vectors and a triangle $A B C$ with side lengths $a, b, c$ satisfy the vector equation
$(20 a - 15 b) \vec{x} + (15 b - 12 c) \vec{y} + (12 c - 20 a) (\vec{x} \times \vec{y}) = \vec{0}$ ,
then $∆ A B C$ is