Find a · → b → if a → = 2 , b → = 5 and a → × b → = 8

Given: a → = 2 , b → = 5 , a → × b → = 8 Let θ be the angle between the two given vectors. Then we know sin θ = a → × b → a → b → Substituting the corresponding values, we get ⇒ sin θ = 8 2 × 5 ⇒ sin θ = 4 5 We also know sin 2 θ + cos 2 θ = 1 ⇒ cos θ = ± 1 - sin 2 θ Substituting the corresponding values, we get ⇒ cos θ = ± 1 - 4 5 2 ⇒ cos θ = ± 1 - 16 25 ⇒ cos θ = ± 9 25 ⇒ cos θ = ± 3 5 Now we know a → · b → = a → b → cos θ Substituting the corresponding values, we get a → · b → = 2 × 5 × ± 3 5 ⇒ a → · b → = ± 6 .

Find the area of the parallelogram whose adjacent sides are a → = i ^ + 2 j ^ + 3 k ^ and b → = 3 i ^ - 2 j ^ + k ^ .

Given a → = i ^ + 2 j ^ + 3 k ^ and b → = 3 i ^ - 2 j ^ + k ^ . Area of the parallelogram with sides a → and b → is given by a → × b → The cross product of these two vectors is a → × b → = i ^ j ^ k ^ 1 2 3 3 - 2 1 ⇒ a → × b → = 2 + 6 i ^ - 1 - 9 j ^ + - 2 - 6 k ^ ⇒ a → × b → = 8 i ^ + 8 j ^ - 8 k ^ And the corresponding magnitude will be ⇒ a → × b → = 8 2 + 8 2 + - 8 2 ⇒ a → × b → = 64 × 3 ⇒ a → × b → = 8 3 Then the area of the parallelogram with these two adjacent sides will be 8 3 sq. units.

Find the area of the parallelogram whose adjacent sides are a → = 3 i ^ + j ^ + 4 k ^ and b → = i ^ - j ^ + k ^

Given a → = 3 i ^ + j ^ + 4 k ^ and b → = i ^ - j ^ + k ^ Area of the parallelogram with sides a → and b → is given by a → × b → The cross product of these two vectors is a → × b → = i ^ j ^ k ^ 3 1 4 1 - 1 1 ⇒ a → × b → = 1 + 4 i ^ - 3 - 4 j ^ + - 3 - 1 k ^ ⇒ a → × b → = 5 i ^ + j ^ - 4 k ^ . And the corresponding magnitude will be a → × b → = 5 2 + 1 2 + - 4 2 ⇒ a → × b → = 25 + 1 + 16 ⇒ a → × b → = 42 . Then the area of the parallelogram with these two adjacent sides will be 42 sq. units.

Find the area of the parallelogram whose adjacent sides are a → = i ^ - j ^ + 3 k ^ and b → = 2 i ^ - 7 j ^ + k ^ .

Given a → = i ^ - j ^ + 3 k ^ and b → = 2 i ^ - 7 j ^ + k ^ . Area of the parallelogram with sides a → and b → is given by a → × b → The cross product of these two vectors is a → × b → = i ^ j ^ k ^ 1 - 1 3 2 - 7 1 , ⇒ a → × b → = - 1 + 21 i ^ - 1 - 6 j ^ + - 7 + 2 k ^ , ⇒ a → × b → = 20 i ^ + 5 j ^ - 5 k ^ . And the corresponding magnitude will be ⇒ a → × b → = 20 2 + 5 2 + - 5 2 , ⇒ a → × b → = 400 + 25 + 25 , ⇒ a → × b → = 450 , ⇒ a → × b → = 15 2 . Then the area of the parallelogram with these two adjacent sides will be 15 2 sq. units.

Find the area of the parallelogram whose diagonals are 2 i ^ - j ^ + k ^ and 3 i ^ + 4 j ^ - k ^ .

Given a → = 2 i ^ - j ^ + k ^ and b → = 3 i ^ + 4 j ^ - k ^ . Area of the parallelogram with diagonals a → and b → is given by 1 2 a → × b → The cross product of these two vectors is a → × b → = i ^ j ^ k ^ 2 - 1 1 3 4 - 1 , ⇒ a → × b → = 1 - 4 i ^ - - 2 - 3 j ^ + 8 + 3 k ^ , ⇒ a → × b → = - 3 i ^ + 5 j ^ + 11 k . ^ And the corresponding magnitude will be ⇒ a → × b → = - 3 2 + 5 2 + 11 2 , ⇒ a → × b → = 9 + 25 + 121 , ⇒ a → × b → = 155 . Then the area of the parallelogram with the given two diagonals will be 1 2 155 sq. units.

Find the area of the parallelogram whose diagonals are the vectors 2 m → - n → and 4 m → - 5 n → , where m → and n → are unit vectors forming an angle of 45 ° .

Given a → = 2 m → - n → and b → = 4 m → - 5 n → . Area of the parallelogram with diagonals p → and q → is given by 1 2 p → × q → The cross product of these two vectors is a → × b → = 2 m → - n → × 4 m → - 5 n → Applying distributive law, we get a → × b → = 2 m → × 4 m → - 2 m → × 5 n → - n → × 4 m → + n → × 5 n → We know, if r is a scalar quantity, then r a → × b → = a → × r b → = r a → × b → , so above equation becomes, a → × b → = 8 m → × m → - 10 m → × n → - 4 n → × m → + 5 n → × n → But we know a → × a → = 0 , so the above equation becomes, a → × b → = 8 0 - 10 m → × n → - 4 n → × m → + 5 0 ⇒ a → × b → = - 10 m → × n → - 4 n → × m → We also know, a → × b → = - b → × a → , so above equation becomes, a → × b → = 10 n → × m → - 4 n → × m → a → × b → = 6 n → × m → And the corresponding magnitude will be ⇒ a → × b → = 6 n → × m → ⇒ a → × b → = 6 n → × m → ⇒ a → × b → = 6 n → m → sin θ But it is given that the angle between the two vectors is 45 ° , so sin 45 ° = 1 2 , substituting this value in above equation, we get ⇒ a → × b → = 6 n → m → × 1 2 ⇒ a → × b → = 6 2 n → m → It is also given that m → and n → are unit vectors, so above equation becomes, ⇒ a → × b → = 6 2 Then the area of the parallelogram with the given two diagonals will be 1 2 × 6 2 = 3 2 sq. units.

Find the vector area of the triangle, the position vectors of whose vertices are i ^ + j ^ + 2 k ^ , 2 i ^ + 2 j ^ - 3 k ^ and 3 i ^ - j ^ - k ^ . Also, find its scalar area.

Given position vectors are A → = i ^ + j ^ + 2 k ^ , B → = 2 i ^ + 2 j ^ - 3 k , ^ C → = 3 i ^ - j ^ - k ^ . So the sides of the triangle will be A B , B C and A C , as shown in the figure. Area of triangle having adjacent sides a → and b → = 1 2 a → × b → . So to find the area, we just need any two adjacent sides. Now we will find the value of two adjacent sides A B → and A C → vectors. ∴ A B → = B → - A → , ⇒ A B → = 2 i ^ + 2 j ^ - 3 k ^ - i ^ + j ^ + 2 k ^ , ⇒ A B → = i ^ + j ^ - 5 k ^ . ∴ A C → = C → - A → , ⇒ A C → = 3 i ^ - j ^ - k ^ - i ^ + j ^ + 2 k ^ , ⇒ A C → = 2 i ^ - 2 j ^ - 3 k . ^ The cross product of these two vectors is A B → × A C → = i ^ j ^ k ^ 1 1 - 5 2 - 2 - 3 , ⇒ A B → × A C → = - 3 - 10 i ^ - - 3 + 10 j ^ + - 2 - 2 k ^ , ⇒ A B → × A C → = - 13 i ^ - 7 j ^ - 4 k ^ . And the corresponding magnitude will be ⇒ A B → × A C → = - 13 2 + - 7 2 + - 4 2 , ⇒ A B → × A C → = 169 + 49 + 16 , ⇒ A B → × A C → = 234 . So the vector area of the given triangle is 1 2 - 13 i ^ - 7 j ^ - 4 k ^ = 1 2 234 . And the scalar area of the triangle with these three vertices will be 234 2 sq. units.

E M B I B E

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Vectors>Chapter 3 - Vector or Cross Product of Two Vectors>EXERCISE

K. C. Sinha and Abhishek Chandra Solutions for Exercise 1: EXERCISE

Author:K. C. Sinha & Abhishek Chandra

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha and Abhishek Chandra Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. VECTOR AND 3-D GEOMETRY for JEE Main and Advanced solutions are prepared by Experienced Embibe Experts.

Questions from K. C. Sinha and Abhishek Chandra Solutions for Exercise 1: EXERCISE with Hints & Solutions

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 3 - Vector or Cross Product of Two Vectors>EXERCISE>Q 3

Find $\vec{a \cdot} \vec{b}$ if $|\vec{a}| = 2, |\vec{b}| = 5$ and $|\vec{a} \times \vec{b}| = 8$

EASY

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 3 - Vector or Cross Product of Two Vectors>EXERCISE>Q 37

Find the area of the parallelogram whose adjacent sides are $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - 2 \hat{j} + \hat{k}$ .

EASY

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 3 - Vector or Cross Product of Two Vectors>EXERCISE>Q 38

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

EASY

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 3 - Vector or Cross Product of Two Vectors>EXERCISE>Q 38

Find the area of the parallelogram whose adjacent sides are $\vec{a} = \hat{i} - \hat{j} + 3 \hat{k}$ and $\vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k}$ .

EASY

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 3 - Vector or Cross Product of Two Vectors>EXERCISE>Q 39

Find the area of the parallelogram whose diagonals are $2 \hat{i} - \hat{j} + \hat{k}$ and $3 \hat{i} + 4 \hat{j} - \hat{k}$ .

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 3 - Vector or Cross Product of Two Vectors>EXERCISE>Q 40

Find the area of the parallelogram whose diagonals are the vectors $2 \vec{m} - \vec{n}$ and $4 \vec{m} - 5 \vec{n}$ , where $\vec{m}$ and $\vec{n}$ are unit vectors forming an angle of $45 °$ .

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 3 - Vector or Cross Product of Two Vectors>EXERCISE>Q 42

Find the vector area of the triangle, the position vectors of whose vertices are $\hat{i} + \hat{j} + 2 \hat{k},$ $2 \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $3 \hat{i} - \hat{j} - \hat{k}$ . Also, find its scalar area.