Let the unit vectors ${\bf{\vec a}}{\rm{\;and\;}}{\bf{\vec b}}$ be perpendicular to each other and the unit vector ${{\bf{\vec c}}}$ be inclined at an angle ${\rm{\theta }}$ to both ${{\bf{\vec a}}}$ and ${\bf{\vec b}}$. If ${\bf{\vec c}} = \alpha ,{\bf{\vec a}} + {\rm{\beta }},{\rm{\;}}{\bf{\vec b}} + \gamma \left( {{\bf{\vec a}}\cdot{\bf{\vec b}}} \right),$ where ${\rm{\alpha }},{\rm{\beta }},{\rm{\gamma }}$ are scalars, then

${\rm{\alpha }} = \cos {\rm{\theta }},{\rm{\;\beta }} = \cos {\rm{\theta }},{\rm{\;}}{{\rm{\gamma }}^2} = \cos 2{\rm{\theta }}$

${\rm{\alpha }} = \cot {\rm{\theta }},{\rm{\;\beta }} = \sin {\rm{\theta }},{\rm{\;}}{{\rm{\gamma }}^2} = \cos 2{\rm{\theta }}$

${\rm{\alpha }} = \cos {\rm{\theta }},{\rm{\;\beta }} = \sin {\rm{\theta }},{\rm{\;}}{{\rm{\gamma }}^2} = \cos 2{\rm{\theta }}$

${\rm{\alpha }} = \sin {\rm{\theta }},{\rm{\;\beta }} = \cos {\rm{\theta }},{\rm{\;}}{{\rm{\gamma }}^2} = \cos 2{\rm{\theta }}$

The angle between the straight lines ${\bf{\vec r}} = \left( {2 - 3t} \right)\widehat {\bf{i}} + \left( {1 + 2t} \right)\widehat {\bf{j}} + \left( {2 + 6t} \right){\bf{\hat k}}$ and ${\bf{\vec r}} = \left( {1 + 4s} \right)\widehat {\bf{i}} + \left( {2 - s} \right)\widehat {\bf{j}} + \left( {8s - 1} \right){\bf{\hat k}}$ is

${\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{34}}{{63}}} \right)$

${\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{\sqrt {41} }}{{34}}} \right)$

${\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{21}}{{34}}} \right)$

${\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{43}}{{63}}} \right)$

Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{{C}}$. Then $\overrightarrow{{C}}+\left(-\frac{1}{2} \hat{{i}}+\frac{1}{3 \sqrt{2}} \hat{{j}}-\frac{\sqrt{2}}{3} \hat{{k}}\right)$ is:

$-\frac{\sqrt{2}}{3} \hat{i}+\frac{\sqrt{2}}{3} \hat{j}+\left(\frac{1}{2}+\frac{2 \sqrt{2}}{3}\right) \hat{{k}}$

$\frac{\sqrt{2}}{3} \hat{{i}}+\frac{1}{3 \sqrt{2}} \hat{{j}}-\frac{1}{2} \hat{{k}}$

$\left(\frac{1}{\sqrt{3}}+\frac{1}{2}\right) \hat{{i}}+\left(\frac{1}{\sqrt{3}}-\frac{1}{3 \sqrt{2}}\right) \hat{{j}}+\left(\frac{1}{\sqrt{3}}+\frac{\sqrt{2}}{3}\right) \hat{{k}}$

If ${\bf{\vec a}},{\bf{\vec b}},{\bf{\vec c}}$ are the pth,qth,nth terms of an HP and ${\rm{\vec u}} = \left( {q - r} \right){\bf{\hat i}} + \left( {r - p} \right){\bf{\hat j}} + \left( {p - q} \right){\bf{\hat k}}$ and ${\rm{\vec v}} = \frac{{{\bf{\hat i}}}}{a} + \frac{{{\bf{\hat j}}}}{b} + \frac{{{\bf{\hat k}}}}{c}$, then

${\bf{\vec u}},\;{\bf{\vec v}}$ are orthogonal vectors

${\bf{\vec u}},\;{\bf{\vec v}}$ are parallel vectors

${\bf{\vec u}}\cdot{\bf{\vec v}} = 1$

${\bf{\vec u}} \times {\bf{\vec v}} = {\bf{\hat i}} + {\bf{\hat j}} + {\bf{\hat k}}$

Let a,b and c be three vectors such that ${\rm{a}}\,{\rm{ = }}\,{\rm{b \times }}\,{\rm{(b \times c)}}{\rm{.}}$ . If magnitude of the vectors a, b and c are $\sqrt 2 ,\,\,1\,\,{\rm{and}}\,\,2$, respectively and the angle between b and c is $\theta \,\,\left( {0 < \theta < \frac{\pi }{2}} \right)\,$, then the Value of $(1 + \tan \theta )$ is equal to

$\frac{{\sqrt 3 + 1}}{{\sqrt 3 }}$

If the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$ be $\theta$ and $\mathbf{a} \cdot \mathbf{b}=\cos \theta$, then the true statement is

$\mathbf{a}$ and $\mathbf{b}$ are unlike vectors

Back to DOT PRODUCT OF VECTORS

Your Progress

0 / 296 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →

Showing 241-270 of 296 questions

Preparing practice and saved progress…

241 Easy

241 of 296

Application +4 | -1

|\vec{a}|=5,|\vec{b}|=4,|\vec{c}|=3

thus what will be the value of

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

, given that

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

View answer and explanation

Correct option: A

Option A (25) is correct because it is the absolute value of the dot product sum, calculated using the condition

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

242 Easy

242 of 296

Application +4 | -1

A tetrahedron has vertices at

O\;\left( {0,\;0} \right),\;A\left( {1,\;2,\;1} \right),\;B\left( {2,\;1,\;3} \right)

and

C\left( { - 1,\;1,\;2} \right)

. Then, the angle between the faces

OAB\;\mathrm{and}\;ABC

will be

View answer and explanation

Correct option: A

{\cos ^{ - 1}}\left( {\frac{{19}}{{35}}} \right)

Option A (

{\cos ^{ - 1}}\left( {\frac{{19}}{{35}}} \right)

) is correct because it represents the angle between the normal vectors of the two faces, which is the angle between the faces.

243 Easy

243 of 296

Analyzing +4 | -1

Let the unit vectors

\mathbf{\vec a}\mathrm{\;and\;}\mathbf{\vec b}

be perpendicular to each other and the unit vector

{\mathbf{\vec c}}

be inclined at an angle

\mathrm{\theta }

to both

{\mathbf{\vec a}}

and

\mathbf{\vec b}

. If

\mathbf{\vec c} = \alpha ,\mathbf{\vec a} + \mathrm{\beta },\mathrm{\;}\mathbf{\vec b} + \gamma \left( {\mathbf{\vec a}\cdot\mathbf{\vec b}} \right),

where

\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }

are scalars, then

View answer and explanation

Correct option: B

\mathrm{\alpha } = \cos \mathrm{\theta },\mathrm{\;\beta } = \cos \mathrm{\theta },\mathrm{\;}{\mathrm{\gamma }^2} = \cos 2\mathrm{\theta }

Option B is correct because by taking dot products of

\vec c

with

\vec a

and

\vec b

, we can find

\alpha

and

\beta

, and then use the magnitude of

\vec c

to find the relation for

\gamma^2

244 Easy

244 of 296

Application +4 | -1

\vec a,\;\,\vec b,\;\,\vec c

are three mutually perpendicular vectors each of magnitude unity, then

\left| {\,\vec a + \vec b + \vec c\,} \right|

is equal to

View answer and explanation

Correct option: C

\sqrt 3

Option C (

\sqrt 3

) is correct because the magnitude of the sum of three mutually perpendicular unit vectors is

\sqrt 3

245 Easy

245 of 296

Application +4 | -1

\mathbf{\vec a},\,\,\mathbf{\vec b}

and

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

are perpendicular to

\mathbf{\vec a} + \mathbf{\vec b}

and

\left| {\,\mathbf{\vec a} + \mathbf{\vec b}\,} \right| = 6

respectively and if

\left| {\,\mathbf{\vec a} + \mathbf{\vec b}\,} \right| = 6

\left| {\,\mathbf{\vec b} + \mathbf{\vec c}\,} \right| = 8

and

\left| {\,\mathbf{\vec c} + \mathbf{\vec a}\,} \right| = 10

, then

\left| {\,\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c}\,} \right|

is equal to

View answer and explanation

Correct option: D

The correct answer is D (10) because by using the given perpendicularity conditions and magnitudes, we can find the magnitude of the sum of the vectors.

246 Easy

246 of 296

Application +4 | -1

A parallelogram is constructed on the vectors

\mathbf{\vec a} = 3\vec \alpha - \mathrm{\vec \beta },\;\mathbf{\vec b} = \mathrm{\vec \alpha } + 3\mathrm{\vec \beta }

, if

\left| {\mathrm{\vec \alpha }} \right| = \left| {\mathrm{\vec \beta }} \right| = 2

and angle between

{\mathrm{\vec \alpha }}

and

\mathrm{\vec \beta }\;\mathrm{is}\frac{\mathrm{\pi }}{3},

then length of diagonal of the parallelogram is

View answer and explanation

Correct option: B

4\sqrt 3

Option A (

4\sqrt{7}

) is correct because it is the length of one of the diagonals of the parallelogram, calculated using vector addition and the given magnitudes and angle.

247 Easy

247 of 296

Analyzing +4 | -1

A vector

\vec{a}

is parallel to the line of intersection of the plane determined by the vectors

\hat{i}, \hat{i}+\hat{j}

and the plane determined by the vectors

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

. The obtuse angle between

\vec{a}

and the vector

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

View answer and explanation

Correct option: A

\frac{3 \pi}{4}

Option A (

\frac{{3 \pi}}{{4}}

) is correct because the vector

\vec{{a}}

is parallel to the intersection of two planes, and the obtuse angle between

\vec{{a}}

and

\vec{{b}}

is found using the dot product formula.

248 Easy

248 of 296

Application +4 | -1

{{\hat n}_1},\;{{\hat n}_2}

are two unit vectors and

\theta

is the angle between them, then

\cos \theta /2 =

View answer and explanation

Correct option: A

\frac{1}{2}\left| {{{\hat n}_1} + {{\hat n}_2}} \right|

The correct answer is A (

\frac{1}{2}\left| {{{\hat n}_1} + {{\hat n}_2}} \right|

) because it is the correct expression for

\cos \theta /2

in terms of the unit vectors

{\hat n}_1

and

{\hat n}_2

and the angle between them.

249 Easy

249 of 296

Application +4 | -1

\left| {\;\mathbf{\vec a}} \right| = 3,\mathrm{\;}\left| {\mathbf{\vec b}} \right| = 4

, then a value of

\lambda

for which

\mathbf{\vec a} + \mathrm{\lambda }\mathbf{\vec b}

is perpendicular to

\mathbf{\vec a} - \mathrm{\lambda }\mathbf{\vec b}

, is

View answer and explanation

Correct option: B

\frac{3}{4}

Option B (

\frac{3}{4}

) is correct because the value of

\lambda

is found by using the perpendicularity condition (dot product equals zero) and the given magnitudes.

250 Easy

250 of 296

Analyzing +4 | -1

If lines

x = ay + b,z = cy + d\mathrm{ }

and

x = a'z + b',y = c'z + d'

are perpendicular, then

View answer and explanation

Correct option: C

aa' + c + c' = 0

Option C (

aa' + c + c' = 0

) is correct because it represents the condition for perpendicularity between two lines in 3D space, derived from the dot product of their direction vectors.

251 Medium

251 of 296

Analyzing +4 | -1

Let

L

be the line of intersection of planes

r.(\hat i - \hat j + 2\hat k) = 2

and

r.(2\hat i + \hat j - \hat k) = 2

. If

P(\alpha ,\,\beta ,\,\gamma )

is the foot of perpendicular on

L

from the point (1, 2, 0), then the value of

35(\alpha + \,\beta + \,\gamma )

is equal to

View answer and explanation

Correct option: B

119

Option B (119) is correct because finding the line of intersection of the two planes, then the foot of the perpendicular from the given point, and finally calculating 35(alpha + beta + gamma) yields 119.

252 Easy

252 of 296

Analyzing +4 | -1

|\mathbf{a}+\mathbf{b}|>|\mathbf{a}-\mathbf{b}|

, then the angle between

\mathbf{a}

and

\mathbf{b}

View answer and explanation

Correct option: A

acute

Option A (acute) is correct because the given inequality implies that the dot product of

\mathbf{a}

and

\mathbf{b}

is positive, which means the angle between them is acute.

253 Easy

253 of 296

Application +4 | -1

\mathbf{a}

and

\mathbf{b}

are two unit vectors such that

\mathbf{a}+2 \mathbf{b}

and

5 a-4 b

are perpendicular to each other, then the angle between

\mathbf{a}

and

\mathbf{b}

View answer and explanation

Correct option: B

60^{\circ}

Option B (

60^{\circ}

) is correct because by using the perpendicularity condition and the fact that a and b are unit vectors, we can find the angle between them.

254 Easy

254 of 296

Application +4 | -1

\mathbf{a}=(1,-1,2), \mathbf{b}=(-2,3,5), \mathbf{c}=(2,-2,4)

and

\mathbf{i}

is the unit vector in the

x

-direction, then

(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}) . \mathbf{i}=

View answer and explanation

Correct option: A

Option A (11) is correct because it is the x-component of the resulting vector (a-2b+3c) after performing the vector operations.

255 Easy

255 of 296

Application +4 | -1

a \mathbf{i}+6 \mathbf{j}-\mathbf{k}

and

7 \mathbf{i}-3 \mathbf{j}+17 \mathbf{k}

are perpendicular vectors, then the value of

a

View answer and explanation

Correct option: A

Option A (5) is correct because the dot product of perpendicular vectors is zero, allowing us to solve for 'a'.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

256 Easy

256 of 296

Analyzing +4 | -1

O A B C

is a tetrahedron such that

\mathrm{OA}^2+\mathrm{BC}^2

=\mathrm{OB}^2+\mathrm{CA}^2=\mathrm{OC}^2+\mathrm{AB}^2

, then which of the following is not true?

View answer and explanation

Correct option: D

A B \perp A C

Option D (

A B \perp A C

) is the correct answer because the given condition implies that

OA \perp BC

OB \perp AC

, and

OC \perp AB

, but not necessarily

AB \perp AC

257 Easy

257 of 296

Analyzing +4 | -1

In a tetrahedron

O A B C

, the edges are of lengths,

|O A|=|B C|=a,|O B|=|A C|=b,|O C|=|A B|=c

. Let

G_1

and

G_2

be the centroids of the triangle

A B C

and

A O C

such that

O G_1 \perp B G_2

, then the value of

\frac{a^2+c^2}{b^2}

View answer and explanation

Correct option: B

The correct answer is B (3) because the given condition

OG_1 \perp BG_2

leads to a relationship between the side lengths of the tetrahedron, resulting in

\frac{a^2+c^2}{b^2}=3

258 Easy

258 of 296

Application +4 | -1

The angle between the straight lines

\mathbf{\vec r} = \left( {2 - 3t} \right)\widehat \mathbf{i} + \left( {1 + 2t} \right)\widehat \mathbf{j} + \left( {2 + 6t} \right)\mathbf{\hat k}

and

\mathbf{\vec r} = \left( {1 + 4s} \right)\widehat \mathbf{i} + \left( {2 - s} \right)\widehat \mathbf{j} + \left( {8s - 1} \right)\mathbf{\hat k}

View answer and explanation

Correct option: D

\mathrm{co}{\mathrm{s}^{ - 1}}\left( {\frac{{34}}{{63}}} \right)

Option D (

\mathrm{co}{\mathrm{s}^{ - 1}}\left( {\frac{{34}}{{63}}} \right)

) is correct because it is the result of applying the dot product formula to find the angle between the direction vectors of the two lines.

259 Easy

259 of 296

Application +4 | -1

If a, b and c are three mutually perpendicular vectors, then the projection of the vectors

l\frac{a}{{|a|}} + m\frac{b}{{|b|}} + n\frac{{(a \times b)}}{{|a \times b|}}

along the angle bisector of the vectors a and b is:

View answer and explanation

Correct option: A

\frac{{l + m}}{{\sqrt 2 }}

Option A (

\frac{{l + m}}{{\sqrt 2 }}

) is correct because it represents the projection of the given vector along the angle bisector of vectors a and b.

260 Medium

260 of 296

Application +4 | -1

Equation of the straight line in the plane

\vec{r}. \vec{n}=d

which is parallel to

\vec{r}=\vec{a}+\lambda \vec{b}

and passes through the foot of perpendicular drawn from the point

P(\vec{a})

to the plane

\vec{r} .\vec{n}=d

is (where

\vec{n} \cdot \vec{b}=0)

View answer and explanation

Correct option: A

\vec{r}=\vec{a}+\left(\frac{d-\vec{a} .\vec{n})}{|\vec{n}|^2}\right)^{\vec{n}+\lambda \vec{b}}

Option A (

\vec{r}=\vec{a}+(\frac{{d-\vec{a}.\vec{n}}}{{|\vec{n}|^2}})^{\vec{n}+\lambda \vec{b}}

) is correct because it represents a line parallel to the given vector and passing through the foot of the perpendicular from point P to the plane.

261 Easy

261 of 296

Application +4 | -1

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

are three vectors such that

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

and

\left| {\,\vec a\,} \right| = 2,\,\;\left| {\,\vec b\,} \right| = 3,\;\,\left| {\,\vec c\,} \right| = 4

, then the value of

\vec a \cdot \vec b\,\, + \,\,\vec b \cdot \vec c\,\, + \,\,\vec c \cdot \vec a

is equal to

View answer and explanation

Correct option: D

- 29/2

Option D (

- 29/2

) is correct because it is the result of expanding the dot product and using the given magnitudes.

262 Easy

262 of 296

Application +4 | -1

Given the vectors

{a}(2,1,-1)

{b}(1,2,1)

{c}(2,-1,3)

{d}(3,-1,2)

, the projection of the vector

{a}+{c}

on the vector

({b}-{d}) \times {c}

View answer and explanation

Correct option: C

\sqrt{6}

Option C (

\sqrt{{6}}

) is correct because it is the result of projecting the vector sum of a and c onto the vector resulting from the cross product of (b-d) and c.

263 Medium

263 of 296

Application +4 | -1

Let a unit vector which makes an angle of

60^{\circ}

with

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

and an angle of

45^{\circ}

with

\hat{i}-\hat{k}

\overrightarrow{{C}}

. Then

\overrightarrow{{C}}+\left(-\frac{1}{2} \hat{{i}}+\frac{1}{3 \sqrt{2}} \hat{{j}}-\frac{\sqrt{2}}{3} \hat{{k}}\right)

is:

View answer and explanation

Correct option: D

\frac{\sqrt{2}}{3} \hat{{i}}-\frac{1}{2} \hat{{k}}

Option D (

\frac{\sqrt{2}}{3} \hat{{i}}-\frac{1}{2} \hat{{k}}

) is correct because, after finding the unit vector

\overrightarrow{{C}}

and adding the given vector, the result matches option D.

264 Medium

264 of 296

Analyzing +4 | -1

Let

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

and

\vec{b}=2 \hat{{i}}-3 \hat{{j}}+5 \widehat{{k}}

. If

\vec{r} \times \vec{a}=\vec{b} \times \vec{r}, \, \vec{r} \cdot(\alpha \hat{{i}}+2 \hat{{j}}+\hat{{k}})=3

and

\vec{r} \cdot(2 \hat{{i}}+5 \hat{{j}}-\alpha \hat{{k}})=-1, \alpha \in R

, then the value of

\alpha+|\vec{r}|^2

is equal to:

View answer and explanation

Correct option: B

Option B (15) is correct because it represents the diameter of the circle, calculated as the distance between the normals plus the diameter of the inner circle.

265 Easy

265 of 296

Application +4 | -1

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

are the pth,qth,nth terms of an HP and

\mathrm{\vec u} = \left( {q - r} \right)\mathbf{\hat i} + \left( {r - p} \right)\mathbf{\hat j} + \left( {p - q} \right)\mathbf{\hat k}

and

\mathrm{\vec v} = \frac{{\mathbf{\hat i}}}{a} + \frac{{\mathbf{\hat j}}}{b} + \frac{{\mathbf{\hat k}}}{c}

, then

View answer and explanation

Correct option: B

\mathbf{\vec u},\;\mathbf{\vec v}

are orthogonal vectors

Option B is correct because the dot product of the vectors

\mathbf{\vec u}

and

\mathbf{\vec v}

is zero, indicating orthogonality.

266 Medium

266 of 296

Application +4 | -1

Let a,b and c be three vectors such that

\mathrm{a}\,\mathrm{ = }\,\mathrm{b \times }\,\mathrm{(b \times c)}\mathrm{.}

. If magnitude of the vectors a, b and c are

\sqrt 2 ,\,\,1\,\,\mathrm{and}\,\,2

, respectively and the angle between b and c is

\theta \,\,\left( {0 < \theta < \frac{\pi }{2}} \right)\,

, then the Value of

(1 + \tan \theta )

is equal to

View answer and explanation

Correct option: B

Option B (2) is correct. The problem involves vector algebra and trigonometric identities, where the angle between vectors b and c is found to be π/4, leading to 1 + tan(θ) = 2.

267 Easy

267 of 296

Application +4 | -1

A vector

\vec{v}

in the first octant is inclined to the

x

-axis at

60^{\circ}

, to the

y

-axis at

45^{\circ}

and to the

z

-axis at an acute angle. If a plane passing through the points

(\sqrt{2},-1,1)

and

(a, b, c)

is normal to

\vec{v}

, then

View answer and explanation

Correct option: B

a+\sqrt{2 b}+c=1

Option B (

a+\sqrt{{2 b}}+c=1

) is correct because the direction cosines of the vector

\vec{{v}}

are found to be

\frac{{1}}{{2}}

\frac{{1}}{{\sqrt{{2}}}}

, and

\frac{{1}}{{2}}

, and using the fact that the plane is normal to

\vec{{v}}

, the equation

a+\sqrt{{2}}b+c=1

is derived.

268 Easy

268 of 296

Application +4 | -1

|\mathbf{a}|=5,|\mathbf{a}-\mathbf{b}|=8

and

|\mathbf{a}+\mathbf{b}|=10

, then

|\mathbf{b}|

is equal to

View answer and explanation

Correct option: B

\sqrt{57}

The correct answer is

\sqrt{57}

because by using the parallelogram law of vector addition and subtraction, we can relate the magnitudes of the vectors and solve for |b|.

269 Easy

269 of 296

Application +4 | -1

The angle between

3 \mathbf{i}+4 \mathbf{j}

and

2 \mathbf{j}-5 \mathbf{k}

View answer and explanation

Correct option: B

\cos ^{-1} \frac{8}{5 \sqrt{29}}

Option D (None of these) is correct because the vectors are not coplanar for any value of m.

270 Easy

270 of 296

Understanding +4 | -1

If the angle between the vectors

\mathbf{a}

and

\mathbf{b}

\theta

and

\mathbf{a} \cdot \mathbf{b}=\cos \theta

, then the true statement is

View answer and explanation

Correct option: D

a and b are unit vectors

Option D (a and b are unit vectors) is correct because if

\mathbf{a} \cdot \mathbf{b} = \cos \theta

, then

|\mathbf{a}||\mathbf{b}| \cos \theta = \cos \theta

, implying

|\mathbf{a}| = |\mathbf{b}| = 1

Revision notes and key takeaways

Review DOT PRODUCT OF VECTORS in your textbook or class notes, then use this set to check your understanding.

Showing 241-270 of 296 questions

Your Progress

Dot product of vectors practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice