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 ^ )

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Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = - \hat{i} - 8 \hat{j} + 2 \hat{k}$ and $\vec{c} = 4 \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ be three vectors such that $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$ . If the angle between the vector $\vec{c}$ and the vector $3 \hat{i} + 4 \hat{j} + \hat{k}$ is $θ$ , then the greatest integer less than or equal to $\tan^{2} θ$ is:

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Important Questions on Vectors

HARD

Mathematics>Coordinate Geometry>Vectors>Cross 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>Vectors>Cross 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

HARD

Mathematics>Coordinate Geometry>Vectors>Cross 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:

MEDIUM

Mathematics>Coordinate Geometry>Vectors>Cross Product of Vectors

A unit vector perpendicular to the plane formed by the points

(1, 0, 1), (0, 2, 2)

and

(3, 3, 0)

EASY

Mathematics>Coordinate Geometry>Vectors>Cross 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>Vectors>Cross Product of Vectors

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

then,

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

EASY

Mathematics>Coordinate Geometry>Vectors>Cross 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:

EASY

Mathematics>Coordinate Geometry>Vectors>Cross 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} =

MEDIUM

Mathematics>Coordinate Geometry>Vectors>Cross 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>Vectors>Cross 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 :

HARD

Mathematics>Coordinate Geometry>Vectors>Cross 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

HARD

Mathematics>Coordinate Geometry>Vectors>Cross 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:

MEDIUM

Mathematics>Coordinate Geometry>Vectors>Cross 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

EASY

Mathematics>Coordinate Geometry>Vectors>Cross 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}

HARD

Mathematics>Coordinate Geometry>Vectors>Cross 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

MEDIUM

Mathematics>Coordinate Geometry>Vectors>Cross Product of Vectors

If the vertices of

Δ ABC

are

A = (2, 3, 5), B = (- 1, 3, 2), C = (3, 5, - 2),

then the area of the

Δ ABC

(in sq. units) is

EASY

Mathematics>Coordinate Geometry>Vectors>Cross Product of Vectors

\bar{O A} = \bar{i} + 2 \bar{j} + 3 \bar{k}

and

\bar{O B} = 4 \bar{i} + \bar{k}

are the position vectors of the points

A

and

B

, then the position vector of a point on the line passing through

B

and parallel to the vector

\bar{O A} \times \bar{O B}

which is at a distance of

\sqrt{189}

units from

B

HARD

Mathematics>Coordinate Geometry>Vectors>Cross 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>Vectors>Cross 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}

MEDIUM

Mathematics>Coordinate Geometry>Vectors>Cross Product of Vectors

Let

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

and

\vec{c}

be a vector such that

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

and

\vec{a} . \vec{c} = 4,

then

{| \vec{c} |}^{2}

is equal to:

Let a→=i^+j^+k^,b→=−i^−8j^+2k^ and c→=4i^+c2j^+c3k^ be three vectors such that b→×a→=c→×a→. If the angle between the vector c→ and the vector 3i^+4j^+k^ is θ, then the greatest integer less than or equal to tan2θ is:

Important Questions on Vectors

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