Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.1 | Set 1

RD Sharma's Class 12 Maths Chapter 24, Scalar or Dot Product Exercise 24.1 solutions are given in this article. These solutions are well designed to help the students in grasping the solving techniques for a wide array of problems. Additionally, they incorporate handy shortcuts and real-world examples to facilitate quick comprehension of concepts and expedite the problem-solving learning curve.

Scalar or Dot Product

Scalar or Dot Product refers to a mathematical operation between two vectors that results in a scalar quantity. It is calculated by multiplying the magnitudes of the vectors and the cosine of the angle between them.

Class 12 Scalar or Dot Product RD Sharma Solution

Question 1. Find \vec{a}.\vec{b} when

(i) \vec{a} = \hat{i}-2 \hat{j}+ \hat{k} and \vec{b} = 4 \hat{i} -4\hat{j} +7\hat{k}

Solution:

 = (\hat{i}-2 \hat{j}+ \hat{k})(\hat{i}-2 \hat{j}+ \hat{k})

= (1)(4) + (-2)(-4) + (1)(7)

= 4 + 8 + 7

= 19

(ii) \vec{a} = \hat{j}+2 \hat{k} and \vec{b} = 2\hat{i}+\hat{k}

Solution:

= (\hat{j}+2 \hat{k} )(2\hat{i}+\hat{k})

= (0)(2) + (1)(0) + (2)(1)

= 2

(iii)  \vec{a} = \hat{j}-\hat{k}  and  \vec{b} = 2\hat{i}+3\hat{j}-2 \hat{k} 

Solution:

= (\hat{j}-\hat{k})(2\hat{i}+3\hat{j}-2 \hat{k} )

= (0)(2) + (1)(3) + (-1)(-2)

= 0 + 3 + 2

= 5

Question 2. For what value of λ are the vector \vec{a}  and \vec{b}  perpendicular to each other? where:

(i) \vec{a} =  λ\hat{i}+2\hat{j}+\hat{k}  and \vec{b} =4\hat{i}-9\hat{j}+2\hat{k}

Solution:

\vec{a}  and \vec{b}  are perpendicular to each other

So \vec{a} . \vec{b} = 0

⇒(λ\hat{i}+2\hat{j}+\hat{k})(4\hat{i}-9\hat{j}+2\hat{k}) = 0         

⇒ λ(4) + (2)(-9) + (1)(2) = 0

⇒ 4λ - 18 + 2 = 0

⇒ 4λ = 16

⇒ λ = 4

(ii) \vec{a} = λ\hat{i}+2\hat{j}+\hat{k}  and \vec{b} =5\hat{i}-9\hat{j}+2\hat{k}

Solution:

\vec{a}  and \vec{b}  are perpendicular to each other

so \vec{a} . \vec{b}  = 0 

⇒ (λ\hat{i}+2\hat{j}+\hat{k})(5\hat{i}-9\hat{j}+2\hat{k})

⇒ λ(5) + (2)(-9) + (1)(2) = 0

⇒ 5λ - 18 + 2 = 0

⇒ 5λ = 16

⇒ λ = 16/5

(iii) \vec{a} = 2\hat{i}+3\hat{j}+4\hat{k}  and \vec{b} =3\hat{i}+2\hat{j}-λ\hat{k}

Solution:

\vec{a}  and \vec{b}  are perpendicular to each other

so \vec{a} . \vec{b}   = 0  

⇒ (2\hat{i}+3\hat{j}+4\hat{k})(3\hat{i}+2\hat{j}-λ\hat{k})   =0

⇒ (2)(3) + (3)(2) - (4)λ = 0

⇒ 6 + 6 - 4λ = 0

⇒ 4λ = 12

⇒ λ = 3

(iv) \vec{a} = λ\hat{i}+3\hat{j}+2\hat{k}  and \vec{b} =\hat{i}-\hat{j}+3\hat{k}

Solution:

\vec{a}  and \vec{b}  are perpendicular to each other

so \vec{a} . \vec{b}=0 

⇒ (λ\hat{i}+3\hat{j}+2\hat{k})(\hat{i}-\hat{j}+3\hat{k})=0  

⇒ λ(1) + (3)(-1) + (2)(3) = 0

⇒ λ - 3 + 6 = 0

⇒ λ = 3

Question 3. If \vec{a}  and  \vec{b}  are two vectors such that |\vec{a}  |=4, |\vec{b}  | = 3 and \vec{a}.\vec{b} = 6. Find the angle between  \vec{a} and  \vec{b}        

Solution:

Let the angle be θ 

cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

= 6 /(4×3) = 1/2

Therefore, θ = cos^-1(1/2)

= π/3

Question 4. If \vec{a} = \hat{i}-\hat{j}  and  \vec{b} =-\hat{j}+2\hat{k}, find (\vec{a}-2\vec{b}). (\vec{a}+\vec{b}) . 

Solution:

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

= \hat{i}-\hat{j}+2\hat{j}-4\hat{k}

= \hat{i}+\hat{j}-4\hat{k}         

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

= \hat{i}-\hat{j}-\hat{j}+2\hat{k}

= \hat{i}-2\hat{j}+2\hat{k}

Now, (\vec{a}-2\vec{b}).(\vec{a}+\vec{b}) 

= (\hat{i}+\hat{j}-4\hat{k})(\hat{i}-2\hat{j}+2\hat{k})

= (1)(1) + (1)(-2) + (-4)(2)

= 1 - 2 - 8

= -9

Therefore, (\vec{a}-2\vec{b}).(\vec{a}+\vec{b}) = -9

Question 5. Find the angle between the vectors \vec{a}  and \vec{b}  where :

(i) \vec{a} = \hat{i}-\hat{j}  and \vec{b} = \hat{j}+\hat{k}

Solution:

 Let the angle be θ between \vec{a}  and \vec{b}

cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

Now, \vec{a} . \vec{b}

= (\hat{i}-\hat{j})(\hat{j}+\hat{k})

= (1)(0) + (-1)(1) + (0)(1)

= 0 - 1 + 0 = -1

|\vec{a}  |= |\hat{i}-\hat{j}  |

= \sqrt{(1)^2+(-1)^2}

= √2

|\vec{b}| = |\hat{j}+\hat{k}|

= \sqrt{(1)^2+(1)^2}

= √2

Now, cos θ = -1/(√2×√2)

= -1/2

θ = cos^-1(-1/2)

= 2π/3

(ii) \vec{a} =3\hat{i}-2\hat{j}-6\hat{k} and \vec{b} =4\hat{i}-\hat{j}+8\hat{k}

Solution:

Let the angle be θ between  \vec{a} and \vec{b}

Now,  \vec{a} . \vec{b}

=(3\hat{i}-2\hat{j}-6\hat{k})(4\hat{i}-\hat{j}+8\hat{k})

=(3)(4) + (-2)(-1) + (-6)(8)

= 12 + 2 - 48

= -34

|\vec{a}| = |3\hat{i}-2\hat{j}-6\hat{k}|

= \sqrt{(3)^2+(-2)^2+(-6)^2}

= √49 = 7

|\vec{b}| = |4\hat{i}-\hat{j}+8\hat{k}|

= \sqrt{(4)^2+(-1)^2+(8)^2}

= √81 = 9

cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

Now, cos θ = -34/(7×9)

= -34/63

θ = cos^-1(-34/63)

(iii) \vec{a} =2\hat{i}-\hat{j}+2\hat{k} and \vec{b} =4\hat{i}+4\hat{j}-2\hat{k}

Solution:

Let the angle be θ between \vec{a} and \vec{b}

Now,  \vec{a} . \vec{b}

=(2\hat{i}-\hat{j}+2\hat{k})(4\hat{i}+4\hat{j}-2\hat{k})

= (2)(4) + (-1)(4) + (2)(-2)

= 8 - 4 - 4 = 0

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

= \sqrt{(2)^2+(-1)^2+(2)^2}

= √9 = 3

|\vec{b}| = |4\hat{i}+4\hat{j}-2\hat{k}|

= \sqrt{(4)^2+(4)^2+(-2)^2}

= √36 = 6

Now, cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}          

cos θ = 0/(3×6) = 0

θ = cos^-1(0)

θ = π/2

(iv) \vec{a} =2\hat{i}-3\hat{j}+\hat{k} and \vec{b} =\hat{i}+\hat{j}-2\hat{k}

Solution:

Let the angle be θ between \vec{a} and \vec{b}

Now, \vec{a} . \vec{b}

=(2\hat{i}-3\hat{j}+\hat{k})(\hat{i}+\hat{j}-2\hat{k})

= (2)(1) + (-3)(1) + (1)(-2)

= 2 - 3 - 2

= -3

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

= \sqrt{(2)^2+(-3)^2+(-1)^2}

= √14 

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

= \sqrt{(1)^2+(1)^2+(-2)^2}

= √6

cos θ =  \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

Now, cos θ = -3/(√14×√6)

= -3/√84

θ = cos^-1(-3/√84)

(v) \vec{a} =\hat{i}+2\hat{j}-\hat{k} and \vec{b} =\hat{i}-\hat{j}+\hat{k}

Solution:

Let the angle be θ between \vec{a} and \vec{b}

Now, \vec{a} . \vec{b}

=(\hat{i}+2\hat{j}-\hat{k})(\hat{i}-\hat{j}+\hat{k})

= (1)(1) + (2)(-1) + (-1)(1)

= 1 - 2 - 1

= -2

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

= \sqrt{(1)^2+(2)^2+(-1)^2}

= √6

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

= \sqrt{(1)^2+(-1)^2+(1)^2}

= √3 

cos θ = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}

Now, cos θ = -2/(√6×√3)

= -2/√18

= -2/3√2

θ = cos^-1(-√2 /3)

Question 6. Find the angles which the vectors \vec{a} =\hat{i}-\hat{j}+\sqrt2\hat{k} makes with the coordinate axes.

Solution:

Components along x, y and z axis are  \hat{i},\hat{j} and \hat{k} respectively.

Let the angle between  \vec{a} and  \hat{i} be θ₁

Now, \vec{a} . \hat{i}

= (\hat{i}-\hat{j}-\sqrt2\hat{k})(\hat{i}-0\hat{j}+0\hat{k})

= (1)(1) + (-1)(0) + (√2)(0) 

= 1

|\vec{a}| = |\hat{i}-\hat{j}+\sqrt2\hat{k}|

= \sqrt{(1)^2+(-1)^2+(√2)^2}

= √4 = 2

|\hat{i}| = |\hat{i}+0\hat{j}+0\hat{k}|

= √1 = 1

cos θ₁ = \frac{\vec{a}.\hat{i}}{|\vec{a}||\hat{i}|}

Now, cos θ₁ = 1/(2×1)

= 1/2

θ₁ = cos^-1(1/2) = π/3

Let the angle between \vec{a} and  \hat{j} be θ₂

Now, \vec{a} . \hat{j}

= (\hat{i}-\hat{j}+\sqrt2\hat{k})(0\hat{i}+\hat{j}+0\hat{k})

= (1)(0) + (-1)(1) + (√2)(0)  

= -1

|\hat{j}| = |0\hat{i}+\hat{j}+0\hat{k}|

= √1 = 1

cos θ₂ = \frac{\vec{a}.\hat{j}}{|\vec{a}||\hat{j}|}

Now, cos θ₂ = -1/(2×1)

= -1/2

θ₂ = cos^-1(-1/2) = 2π/3

 Let the angle between \vec{a} and \hat{k} be θ₃

Now, \vec{a} . \hat{k}

= (\hat{i}-\hat{j}+\sqrt2\hat{k})(0\hat{i}+0\hat{j}+\hat{k})

= (1)(0) + (-1)(0) + (√2)(1)  

= √2

|\hat{k}| = |0\hat{i}+0\hat{j}+\hat{k}|

= √1 = 1

cos θ₃ =  \frac{\vec{a}.\hat{k}}{|\vec{a}||\hat{k}|}

= 1/(√2)

= cos^-1(1/√2) = π/4

Question 7(i). Dot product of a vector with \hat{i}+\hat{j}-3\hat{k}, \hat{i}+3\hat{j}-2\hat{k} and 2\hat{i}+\hat{j}+4\hat{k} are 0, 5 and 8respectively. Find the vector.

Solution:

Let \vec{a} =\hat{i}+\hat{j}-3\hat{k}, \vec{b} =\hat{i}+3\hat{j}-2\hat{k} and  \vec{c}=2\hat{i}+\hat{j}+4\hat{k} be three given vectors.

Let \vec{r}=x\hat{i}+y\hat{j}+j\hat{k} be a vector such that its dot products with  \vec{a},  \vec{b}, and \vec{c} are 0, 5 and 8 respectively. Then, 

 \vec{r}. \vec{a} = 0

⇒ (x\hat{i}+y\hat{j}+j\hat{k})(\hat{i}+\hat{j}-3\hat{k})       = 0

⇒ x + y - 3z = 0        ....(1)

 \vec{r}. \vec{b} = 5

⇒ (x\hat{i}+y\hat{j}+j\hat{k})(\hat{i}+3\hat{j}-2\hat{k})       = 5

⇒ x + 3y - 2z = 5     .....(2)

 \vec{r}. \vec{c} = 8

⇒ (x\hat{i}+y\hat{j}+j\hat{k})(2\hat{i}+\hat{j}+4\hat{k})       = 8

⇒ 2x + y + 4z = 8    .....(3)

Solving 1,2 and 3 we get x = 1, y = 2 and z = 1,

Hence, the required vector is \vec{r}=\hat{i}+2\hat{j}+\hat{k}

Question 8. If  \vec{a} and  \vec{b} are unit vectors inclined at an angle θ then prove that 

(i) cos θ/2 = 1/2|\hat{a}+\hat{b}|

Solution:

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

|\hat{a}+\hat{b}|² =(\hat{a}+\hat{b})² 

= (\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b}

= 1 + 1 + 2\hat{a}.\hat{b} 

= 2 + 2|\hat{a}||\hat{b}|cos θ 

= 2(1 + (1)(1)cos θ)

= 2(2cos² θ/2)

|\hat{a}+\hat{b}|² = 4cos² θ/2

\hat{a}+\hat{b} = 2 cos θ/2

cos θ/2 = 1/2|\hat{a}+\hat{b}|

(ii) tan θ/2 = \frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|} 

Solution:

|\hat{a}| = |\hat{b}| = 1

\frac{|\hat{a}-\hat{b}|^2}{|\hat{a}+\hat{b}|^2}= \frac{(\hat{a}-\hat{b})^2}{(\hat{a}+\hat{b})^2} 

=\frac{(\hat{a})^2+(\hat{b})^2-2\hat{a}.\hat{b}}{(\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b}}

= \frac{2-2|\hat{a}||\hat{b}|cos θ }{2+2|\hat{a}||\hat{b}|cos θ }

=\frac{2(1-cos θ)}{2(1+cos θ)}

= \frac{2sin^2 θ/2}{2cos^2 θ/2}

= tan² θ/2

Therefore, tan θ/2 =\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|} 

Question 9. If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is √3.

Solution:

Let \hat{a} and \hat{b} be two unit vectors

Then,  |\hat{a}| = |\hat{b}| = 1

According to question:

|\hat{a}+\hat{b}| = 1

Taking square on both sides

⇒|\hat{a}+\hat{b}|^2 = (1)^2

⇒(\hat{a})^2+(\hat{b})^2+2\hat{a}.\hat{b} = 1

⇒ (1)²+(1)²+2\hat{a}.\hat{b} = 1

⇒ 2+ 2\hat{a}.\hat{b} = 1

⇒ 2\hat{a}.\hat{b}= -1

⇒ \hat{a}.\hat{b} =-1/2

Now, |\hat{a}-\hat{b}|^2 = (\hat{a}-\hat{b})^2

= (\hat{a})^2 + (\hat{b})^2 - 2\hat{a}.\hat{b}

= (1)² + (1)²  - 2 (-1/2)

= 2 + 1 = 3

Therefore, |\hat{a}-\hat{b}|^2 = 3 

|\hat{a}-\hat{b}|=√3

Question 10. If \vec{a},\vec{b},\vec{c}  are three mutually perpendicular unit vectors, then prove that |\vec{a}+\vec{b}+\vec{c} | =√3.

Solution:

Given \vec{a},\vec{b},\vec{c}   are mutually perpendicular so,

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

|\vec{a}| = |\vec{b}| = |\vec{c}|=1      

Now, 

|\vec{a}+\vec{b}+\vec{c}|^2   = (\vec{a}+\vec{b}+\vec{c})^2

=(\vec{a})^2+(\vec{b})^2+(\vec{c})^2+2\vec{a}\vec{b}+2\vec{b}\vec{c}+2\vec{c}\vec{a}

= (1)² + (1)² +(1)² + 0

= 3

 |\vec{a}+\vec{b}+\vec{c}|   = √3

Question 11. If |\vec{a}+\vec{b}| = 60, |\vec{a}-\vec{b}| = 40 and |\vec{b}|= 46, find |\vec{a}|

Solution:

Given |\vec{a}+\vec{b}|=60, |\vec{a}-\vec{b}| = 40 and  |\vec{b}|= 46

We know that, 

(a + b)² + (a - b)² = 2(a² + b²)

⇒ |\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2 = 2(|\vec{a}|^2+|\vec{b}|^2)

⇒ 60² + 40² = 2(|\vec{a}| ² + 49²)

⇒ 3600 + 1600 = 2|\vec{a}|^2   + 2401 

⇒ 2|\vec{a}|   = 968

⇒ |\vec{a}|   = √484 =22

Question 12. Show that the vector \hat{i}+\hat{j}+\hat{k} is equally inclined with the coordinate axes. 

Solution:

Let \vec{a} = \hat{i}+\hat{j}+\hat{k}

|\vec{a}| =√(1+1+1) = √3

Let θ₁, θ₂, θ₃ be the angle between the coordinate axes and the \vec{a}

cos θ₁ = \frac{\vec{a}.\hat{i}}{|\vec{a}||\hat{i}|}

= 1/√3

cos θ₂ = \frac{\vec{a}.\hat{j}}{|\vec{a}||\hat{j}|}

= 1/√3

cos θ₃ = \frac{\vec{a}.\hat{k}}{|\vec{a}||\hat{k}|}

= 1/√3

Since, cos θ₁ = cos θ₂ = cos θ₃ 

Therefore, Given vector is equally inclined with coordinate axis.

Question 13. Show that the vectors \vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}), \vec{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\vec{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k}) are mutually perpendicular unit vectors.

Solution: 

Given, \vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})

\vec{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})

\vec{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k}) 

|\vec{a}|   = (1/7)√(2² + 3² + 6²) = (1/7)(√49) = 1

|\vec{b}|   = (1/7)√(3² + (-6)² + 2²) = (1/7)(√49) = 1

|\vec{c}|   = (1/7)√(6² + 2² + (-3)²) = (1/7)(√49) = 1

Now, \vec{a}.\vec{b} =   1/49[3 × 2 - 3 × 6 + 6 × 2]

= 1/49[6 - 18 + 12] = 0 

\vec{b}.\vec{c} =   1/49[3 × 6 - 6 × 2 - 2 × 3]

= 1/49[18 - 12 - 6] = 0

Since, \vec{a}.\vec{b} = \vec{b}.\vec{c}=0 they are mutually perpendicular unit vectors.

Question 14. For any two vectors \vec{a}  and \vec{b}  , Show that (\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0\Leftrightarrow|\vec{a}|=|\vec{b}|.

Solution:

To prove (\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0\Leftrightarrow|\vec{a}|=|\vec{b}|

⇒(\vec{a}+\vec{b}).(\vec{a}-\vec{b})=0

⇒|\vec{a}|^2-|\vec{b}|^2=0

⇒|\vec{a}|=|\vec{b}|    

Hence Proved

Question 15. If \vec{a}=2\hat{i}-\hat{j}+\hat{k} , \vec{b}=\hat{i}+\hat{j}-2\hat{k} and \vec{c}=\hat{i}+3\hat{j}-\hat{k} , find such that \vec{a} is perpendicular to λ\vec{b}+\vec{c} .

Solution:

Given: \vec{a}=2\hat{i}-\hat{j}+\hat{k}

\vec{b}=\hat{i}+\hat{j}-2\hat{k}  

\vec{c}=\hat{i}+3\hat{j}-\hat{k}

According to question

\vec{a}(λ\vec{b}+\vec{c})=0

⇒ (2\hat{i}-\hat{j}+\hat{k})[λ(\hat{i}+\hat{j}-2\hat{k})+(\hat{i}+3\hat{j}-\hat{k})]=0

⇒ (2\hat{i}-\hat{j}+\hat{k})(λ\hat{i}+λ\hat{j}-2λ\hat{k}+\hat{i}+3\hat{j}-\hat{k})=0

⇒ 2(λ+1) - (λ+3) -2λ-1 = 0

⇒ 2λ + 2 -λ - 3 - 2λ - 1 = 0

⇒ -λ = 2

⇒ λ = -2

Question 16. If \vec{p}=5\hat{i}+λ\hat{j}-3\hat{k} and \vec{q}=\hat{i}+3\hat{j}-5\hat{k} , then find the value of λ so that \vec{p}+\vec{q}  and \vec{p}-\vec{q}  are perpendicular vectors.

Solution:

Given, \vec{p}=5\hat{i}+λ\hat{j}-3\hat{k}   

\vec{q}=\hat{i}+3\hat{j}-5\hat{k}

According to question

(\vec{p}+\vec{q})(\vec{p}-\vec{q})=0

⇒|\vec{p}|^2-|\vec{q}|^2=0  

⇒ |\vec{p}|^2=|\vec{q}|^2

⇒ \sqrt{5^2+λ^2+(-3)^2}=\sqrt{1^2+3^2+(-5)^2}

⇒ 25 + λ² + 9 = 1 + 9 + 25

⇒ λ² = 1

⇒ λ = 1

Practice Problems:

1. If a = 3i - 2j + k and b = i + 4j - 2k, find a · b.

2. Determine if vectors a = 2i + j - 3k and b = i - 2j + k are perpendicular.

3. Calculate the angle between vectors a = i + j + k and b = i - j + k.

4. Find the magnitude of vector a if a · a = 25.

5. If a = 2i + 3j - k and b = i - j + 2k, find |a + b|.

6. Prove that a · (b × c) = b · (c × a) = c · (a × b).

7. Find a unit vector in the direction of a = 3i - 4j + 12k.

8. If a · b = 5, b · c = -2, and c · a = 10, find (a + b) · (b + c).

9. Determine the value of k if (3i + 2j + k) · (i - j + 2k) = 10.

10. Find the projection of a = 2i + 3j - k onto b = i + j + k.

Summary

Chapter 24 of RD Sharma Class 12 Solutions covers the Scalar or Dot Product of vectors. This chapter introduces the concept of scalar product and its properties. Key topics include:

• Definition and calculation of scalar product
• Geometric interpretation of scalar product
• Properties of scalar product (commutative, distributive over addition)
• Applications in finding vector magnitudes and angles
• Projection of one vector onto another
• Conditions for perpendicularity and parallelism
• Vector equations involving scalar product

The chapter emphasizes the importance of scalar product in vector algebra and its applications in physics and mathematics.

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