Find the angle between the vectors 3 i ^ + j ^ + 2 k ^ and 2 i ^ - 2 j ^ + 4 k ^ .

Given that, the vectors are 3 i ^ + j ^ + 2 k ^ and 2 i ^ - 2 j ^ + 4 k ^ . We know that, cos θ = a ⇀ · b ⇀ a ⇀ b ⇀ where a ⇀ and b ⇀ are two vectors and the angle between them is θ . ⇒ cos θ = 3 i ^ + j ^ + 2 k ^ · 2 i ^ - 2 j ^ + 4 k ^ 3 i ^ + j ^ + 2 k ^ 2 i ^ - 2 j ^ + 4 k ^ ⇒ cos θ = 6 - 2 + 8 3 2 + 1 2 + 2 2 2 2 + - 2 2 + 4 2 ⇒ cos θ = 12 14 24 = 3 7 3 = 3 7 ⇒ cos θ = 3 7 ⇒ θ = 49 . 106 °

Find the angle between the vectors 3 i ^ + j ^ + 2 k ^ and 2 i ^ - 2 j ^ + 4 k ^ .

Given that, the vectors are 3 i ^ + j ^ + 2 k ^ and 2 i ^ - 2 j ^ + 4 k ^ . We know that, cos θ = a ⇀ · b ⇀ a ⇀ b ⇀ where a ⇀ and b ⇀ are two vectors and the angle between them is θ . ⇒ cos θ = 3 i ^ + j ^ + 2 k ^ · 2 i ^ - 2 j ^ + 4 k ^ 3 i ^ + j ^ + 2 k ^ 2 i ^ - 2 j ^ + 4 k ^ ⇒ cos θ = 6 - 2 + 8 3 2 + 1 2 + 2 2 2 2 + - 2 2 + 4 2 ⇒ cos θ = 12 14 24 = 3 7 3 = 3 7 ⇒ cos θ = 3 7 ⇒ θ = 49 . 106 °

If G is the centroid of a triangle A B C , prove that G A → + G B → + G C → = 0 → .

For any triangle A B C , Given G A → + G B → + G C → = 0 → . Now G is the centroid of triangle A B C , which divides the medians A D , B E and C F in the ratio 2 : 1 ⇒ A G → : G D → = 2 : 1 ⇒ G A → = - 2 3 A D → Similarly, G B → = - 2 3 B E → and G C → = - 2 3 C F → Now, LHS = G A → + G B → + G C → = - 2 3 A D → + B E → + C F → = - 2 3 A B → + 1 2 B C → + B C → + 1 2 C A → + C A → + 1 2 A B → = - 2 3 3 2 ( A B → + B C → + C A → ) = - 1 0 → = 0 → = RHS .

If a → = 3 i ^ + 2 j ^ + 2 k ^ , b → = - i ^ + 3 j ^ - k ^ and c → = i ^ + j ^ + k ^ then find a → × ( b → × c → ) .

Given that, the vectors are a → = 3 i ^ + 2 j ^ + 2 k ^ , b → = - i ^ + 3 j ^ - k ^ and c → = i ^ + j ^ + k ^ . We know that, a → × ( b → × c → ) = ( a → · c → ) b → - ( a → · b → ) c → . ⇒ a → × ( b → × c → ) = 3 i ^ + 2 j ^ + 2 k ^ · i ^ + j ^ + k ^ - i ^ + 3 j ^ - k ^ - 3 i ^ + 2 j ^ + 2 k ^ · - i ^ + 3 j ^ - k ^ i ^ + j ^ + k ^ = 8 - i ^ + 3 j ^ - k ^ - 1 i ^ + j ^ + k ^ = - 8 i ^ + 24 j ^ - 8 k ^ - i ^ + j ^ + k ^ = - 9 i ^ + 23 j ^ - 9 k ^

Find a unit vector in the direction of the vector a → = 2 i ^ + 3 j ^ + k ^ .

In order to find a unit vector in the direction of a given vector, we have to divide that vector by its magnitude. The magnitude of vector a → = 2 i ^ + 3 j ^ + k ^ can be given as, a → = 2 2 + 3 2 + 1 2 ⇒ a → = 4 + 9 + 1 ⇒ a → = 14 Therefore, the unit vector in the direction of a → (assume u a → ) can be given as, u a → = a → a → ⇒ u a → = 1 14 2 i ^ + 3 j ^ + k ^ ⇒ u a → = 2 14 i ^ + 3 14 j ^ + 1 14 k ^

Find the projection of a → = 2 i ^ + 3 j ^ + 2 k ^ on the vector b → = i ^ + 2 j ^ + k ^ .

The projection of a vector a → on b → is given by a → · b → b → . Therefore, a → · b → b → = 2 i ^ + 3 j ^ + 2 k ^ · i ^ + 2 j ^ + k ^ 1 2 + 2 2 + 1 2 = 2 × 1 + 3 × 2 + 2 × 1 1 + 4 + 1 = 2 + 6 + 2 6 = 10 6

If a → + b → + c → = 0 → , then prove that a → × b → = b → × c → = c → × a → .

Given that, a → + b → + c → = 0 → and To prove: a → × b → = b → × c → = c → × a → Here, a → = - ( b → + c → ) ⇒ a → × b → = - ( b → + c → ) × b → = - b → × b → - c → × b → = 0 → + b → × c → = b → × c → [Using a → × a → = b → × b → = c → × c → = 0 ] Again, b → = - ( a → + c → ) ⇒ a → × b → = - a → × ( a → + c → ) = - a → × a → - a → × c → = 0 - a → × c → = c → × a → Thus, a → × b → = b → × c → = c → × a → Hence, proved.

Find a unit vector perpendicular to each of the vectors a → + b → and a → - b → , where a → = i ^ + j ^ + k ^ and b → = i ^ + 2 j ^ + 3 k ^

The given two vectors are a → = i ^ + j ^ + k ^ and b → = i ^ + 2 j ^ + 3 k ^ . Now we need to find a → + b → and a → - b → . So, a → + b → = i ^ + j ^ + k ^ + i ^ + 2 j ^ + 3 k ^ a → + b → = 2 i ^ + 3 j ^ + 4 k ^ And, a → - b → = i ^ + j ^ + k ^ - i ^ + 2 j ^ + 3 k ^ a → - b → = - j ^ - 2 k ^ Now the cross product of these vectors will be a → + b → × a → - b → = i ^ j ^ k ^ 2 3 4 0 - 1 - 2 ⇒ a → + b → × a → - b → = - 2 i ^ + 4 j ^ - 2 k ^ And the corresponding magnitude will be a → + b → × a → - b → = ( - 2 ) 2 + ( 4 ) 2 + ( - 2 ) 2 ⇒ a → + b → × a → - b → = 4 + 16 + 4 ⇒ a → + b → × a → - b → = 24 ⇒ a → + b → × a → - b → = 6 × 4 ⇒ a → + b → × a → - b → = 2 6 Now we know vector the product a → + b → × a → - b → will be perpendicular to the plane of a → + b → and a → - b → . So the unit vector perpendicular to the plane of a → + b → and a → - b → is given as = a → + b → × a → - b → a → + b → × a → - b → Now substituting the corresponding value, we get = - 2 i ^ + 4 j ^ - 2 k ^ 2 6 = - i ^ + 2 j ^ - k ^ 6 Or, the unit vector can be written as = - 1 6 i ^ + 2 6 j ^ - 1 6 k ^ .

Show that three points A ( - 2 i ^ + 3 j ^ + 5 k ^ ) , B ( i ^ + 2 j ^ + 3 k ^ ) and C ( 7 i ^ - k ^ ) are collinear.

A B → = ( i ^ + 2 j ^ + 3 k ^ ) - ( - 2 i ^ + 3 j ^ + 5 k ^ ) = 3 i ^ - j ^ - 2 k ^ B C → = ( 7 i ^ - k ^ ) - ( i ^ + 2 j ^ + 3 k ^ ) = 6 i ^ - 2 j ^ - 4 k ^ Now , B C → = 6 i ^ - 2 j ^ - 4 k ^ = 2 ( 3 i ^ - j ^ - 2 k ^ ) B C → = λ A B → , where λ = 2 ∴ B C → and A B → are collinear Since , point B is common to both , A , B and C are collinear .

Find a vector in the direction of the vector a → = i ^ - 2 j ^ that has magnitude of 7 units.

To find a vector in the direction of vector a → with a magnitude of 7 units, we first need to find the unit vector in the direction of vector a → . The unit vector is given as, u a → = a → a → ⇒ u a → = i ^ - 2 j ^ 1 2 + - 2 2 ⇒ u a → = i ^ - 2 j ^ 1 + 4 ⇒ u a → = i ^ - 2 j ^ 5 ⇒ u a → = 1 5 i ^ - 2 5 j ^ Now, we have got a unit vector in the direction of a → . The vector of a magnitude of 7 units in the direction of a → can be obtained as, 7 u a → = 7 1 5 i ^ - 2 5 j ^ ⇒ 7 u a → = 7 5 i ^ - 14 5 j ^

Find the angle between two vectors a → and b → with magnitudes 1 and 2 , respectively, and satisfying a → · b → = 1

The angle between two vectors a → and b → can be find out using the formula, cos θ = a → · b → a → b → In the question, a → · b → = 1 , a → = 1 and b → = 2 is given, so substituting these values, cos θ = 1 1 × 2 cos θ = 1 2 θ = cos - 1 1 2 θ = 60 ° or π 3 .

If a → + b → + c → = 0 → , then prove that a → × b → = b → × c → = c → × a → .

Given that, a → + b → + c → = 0 → and To prove: a → × b → = b → × c → = c → × a → Here, a → = - ( b → + c → ) ⇒ a → × b → = - ( b → + c → ) × b → = - b → × b → - c → × b → = 0 → + b → × c → = b → × c → [Using a → × a → = b → × b → = c → × c → = 0 ] Again, b → = - ( a → + c → ) ⇒ a → × b → = - a → × ( a → + c → ) = - a → × a → - a → × c → = 0 - a → × c → = c → × a → Thus, a → × b → = b → × c → = c → × a → Hence, proved.

Find the projection of a → = 2 i ^ + 3 j ^ + 2 k ^ on the vector b → = i ^ + 2 j ^ + k ^ .

The projection of a vector a → on b → is given by a → · b → b → . Therefore, a → · b → b → = 2 i ^ + 3 j ^ + 2 k ^ · i ^ + 2 j ^ + k ^ 1 2 + 2 2 + 1 2 = 2 × 1 + 3 × 2 + 2 × 1 1 + 4 + 1 = 2 + 6 + 2 6 = 10 6

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2018>Q 10

EASY

12th Madhya Pradesh Board

IMPORTANT

Earn 100

Find the angle between the vectors $3 \hat{i} + \hat{j} + 2 \hat{k} and 2 \hat{i} - 2 \hat{j} + 4 \hat{k}$ .

Important Questions on Vector Algebra

HARD

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2018>Q 11

If

\Rightarrow \cos θ = \sqrt{\frac{3}{7}}

is the centroid of a triangle

A B C

, prove that

\vec{G A} + \vec{G B} + \vec{G C} = \vec{0}

.

EASY

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2018>Q 12

If

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

and

\vec{c} = \hat{i} + \hat{j} + \hat{k}

then find

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

.

EASY

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2019>Q 13

Find a unit vector in the direction of the vector

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

.

MEDIUM

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2019>Q 14

Find the projection of

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

on the vector

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

.

MEDIUM

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2019>Q 15

If

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

, then prove that

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

.

MEDIUM

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2019>Q 16

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a} = 3 \hat{i} + \hat{j} + 4 \hat{k} and \vec{b} = \hat{i} - \hat{j} + \hat{k}$ .