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

Question 17. If \vec{α}=3\hat{i}+4\hat{j}+5\hat{k} and \vec{β}=2\hat{i}+\hat{j}-4\hat{k} , then express \vec{β} in the form of \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} .

Solution:

Given, \vec{α}=3\hat{i}+4\hat{j}+5\hat{k}
\vec{β}=2\hat{i}+\hat{j}-4\hat{k}
According to question
\vec{β_1} = λ\vec{α} also \vec{β_2}.\vec{α} = 0
Now,
\vec{β_1} = λ(3\hat{i}+4\hat{j}+5\hat{k})
⇒ \vec{β_1} = 3λ\hat{i}+4λ\hat{j}+5λ\hat{k}
\vec{β_2}=\vec{β}-\vec{β_1}
⇒ \vec{β_2}=2\hat{i}+\hat{j}-4\hat{k}-(3λ\hat{i}+4λ\hat{j}+5λ\hat{k})
⇒ \vec{β_2}=(2-3λ)\hat{i}+(1-4λ)\hat{j}-(4+5λ)\hat{k}
Now,
\vec{β_2}.\vec{α}=0
⇒ [(2-3λ)\hat{i}+(1-4λ)\hat{j}-(4+5λ)\hat{k}](3\hat{i}+4\hat{j}+5\hat{k})=0
⇒ 3(2-3λ)+4(1-4λ)-5(4+5λ) = 0
⇒ 6-9λ+4-16λ-20-25λ = 0
⇒ -10 -50λ = 0
⇒ λ = -1/5
\vec{β_1} = -3/5\hat{i}-4/5\hat{j}-1\hat{k}
\vec{β_2} = 13/5\hat{i}+9/5\hat{j}-3\hat{k}

Question 18. If either \vec{a}=\vec{0} or \vec{b}=\vec{0} , then \vec{a}.\vec{b}=0 . But, The converse need not be true. Justify your answer with an example.

Solution:

Given,
\vec{a}=\vec{0}    or \vec{b}=\vec{0}    then \vec{a}.\vec{b}=0
Suppose \vec{a}=2\hat{i}+\hat{j}+\hat{k}
\vec{b}=\hat{i}-\hat{j}-\hat{k}
\vec{a}.\vec{b}=0
But,
|\vec{a}|   = √(2)²+(1)²+(1)²
= √4+1+1
= √6 ≠ 0
|\vec{b}|   = √(1)²+(1)²+(1)²
= √3 ≠ 0
Hence Proved

Question 19. Show that the vectors \vec{a}=3\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}-3\hat{j}+5\hat{k},\vec{c}=2\hat{i}+\hat{j}-4\hat{k} form a right-angled triangle.

Solution:

Given, \vec{a}=3\hat{i}-2\hat{j}+\hat{k}
\vec{b}=\hat{i}-3\hat{j}+5\hat{k}
\vec{c}=2\hat{i}+\hat{j}-4\hat{k}
To prove given vectors form a right angle triangle
|\vec{a}|  = √(3²+(-2)²+1²) = √14
|\vec{b}|  = √(1²+(-3)²+5²) = √35
|\vec{c}|  = √(2²+1²+(-4)²) = √21
|\vec{a}|^2+|\vec{c}|^2 = (\sqrt14)^2+(\sqrt21)^2
= 14 + 21 = 35
Since, |\vec{b}|^2=|\vec{a}|^2+|\vec{c}|^2       (Pythagoras Theorem)
Hence, \vec{a},\vec{b}  and \vec{c}   form a right angled triangle.

Question 20. If \vec{a}=2\hat{i}+2\hat{j}+3\hat{k} , \vec{b}=-\hat{i}+2\hat{j}+\hat{k} and \vec{c}=3\hat{i}+\hat{j} are such that \vec{a}+λ\vec{b} is perpendicular to \vec{c} , then find the value of λ.

Solution:

Given:
\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}
\vec{b}=-\hat{i}+2\hat{j}+\hat{k}
\vec{c}=3\hat{i}+\hat{j}
Now, (\vec{a}+λ\vec{b}).\vec{c}=0
⇒ [2\hat{i}+2\hat{j}+3\hat{k}+λ(-\hat{i}+2\hat{j}+\hat{k})](3\hat{i}+\hat{j})=0
⇒ [(2-λ)\hat{i}+(2+2λ)\hat{j}+(3+λ)\hat{k}](3\hat{i}+\hat{j})=0
⇒ (2 - λ)3 + (2 + 2λ) + 0 = 0
⇒ 6 - 3λ + 2 + 2λ =0
⇒ λ = 8

Question 21. Find the angles of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4), and C(5, 7, 1).

Solution:

Given that angle of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4) and C(5, 7, 1).
\vec{AB}=(\vec{B}-\vec{A}) =3\hat{i}+2\hat{j}+6\hat{k}
\vec{BC}=(\vec{C}-\vec{B}) =2\hat{i}+6\hat{j}-3\hat{k}
\vec{AC}=(\vec{C}-\vec{A}) =5\hat{i}+8\hat{j}+3\hat{k}
|\vec{AB}|=\sqrt{3^2+2^2+6^2}=7
|\vec{BC}|=\sqrt{2^2+6^2+(-3)^2}=7
|\vec{AC}|=\sqrt{5^2+8^2+3^2}
= √98 = 7√2
Now,
\vec{AB}.\vec{BC}   = (3 × 2 + 2 × 6 - 6 × 3) = 0
Thus, we can say AB is perpendicular to BC.
Hence, AB = BC = 7, ∠A =∠C and ∠B = 90°
∠A + ∠B + ∠C = 180°
2∠A = 180° - 90°
∠A = 45°
∠C = 45°
∠B = 90°

Question 22. Find the magnitude of two vectors \vec{a} and \vec{b} , having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.

Solution:

We know \vec{a}.\vec{b}=|\vec{a}||\vec{b}|cos θ
⇒ 1/ 2 = |\vec{a}||\vec{a}|cos θ
⇒ 1/2 = |\vec{a}|^2 (1/2)
⇒ |\vec{a}| = 1
or
⇒ |\vec{a}| = |\vec{b}| = 1

Question 23. Show that the points whose position vector are \vec{a} =4\hat{i}-3\hat{j}+\hat{k},\vec{b} =2\hat{i}-4\hat{j}+5\hat{k}, \vec{c} =\hat{i}-\hat{j} form a right triangle.

Solution:

Given that positions vectors
\vec{a} =4\hat{i}-3\hat{j}+\hat{k}
\vec{b} =2\hat{i}-4\hat{j}+5\hat{k}
\vec{c} =\hat{i}-\hat{j}
Now,
\vec{AB} = (\vec{b}-\vec{a})
⇒ \vec{AB} =-2\hat{i}-\hat{j}+4\hat{k}
\vec{BC} = (\vec{c}-\vec{b})
⇒ \vec{BC} =-\hat{i}+3\hat{j}-5\hat{k}
\vec{CA} = (\vec{a}-\vec{c})
⇒ \vec{CA} =3\hat{i}-2\hat{j}+\hat{k}
Now, \vec{AB}.\vec{BC} =(-2\hat{i}-\hat{j}+4\hat{k})(-\hat{i}+3\hat{j}-5\hat{k})
= 2 - 3 - 20 = -21
\vec{BC}.\vec{CA} =(-\hat{i}+3\hat{j}-5\hat{k})(3\hat{i}-2\hat{j}+\hat{k})
= -3 - 6 - 5 = -14
\vec{AB}.\vec{CA} =(-2\hat{i}-\hat{j}+4\hat{k})(3\hat{i}-2\hat{j}+\hat{k})
= -6 + 2 + 4 = 0
So, AB is perpendicular to CA or the given position vectors form a right-angled triangle.

Question 24. If the vertices A, B, C of △ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, what is the magnitude of ∠ABC?

Solution:

Given the vertices of △ABC are A(1, 2, 3), B(-1, 0, 0), C(0, 1, 2)
Now, \vec{AB} = (-1-1)\hat{i}+(0-2)\hat{j}+(0-3)\hat{k}
= -2\hat{i}-2\hat{j}-3\hat{k}
Or, \vec{BA}=2\hat{i}+2\hat{j}+3\hat{k}
\vec{BC}=\hat{i}+\hat{j}+2\hat{k}
We know that \vec{BA}.\vec{BC} = |\vec{BA}||\vec{BC}|
\vec{BA}.\vec{BC} =   (2 × 1) + (2 × 1) + (3 × 2)
= 2 + 2 + 6 = 10
Now, |\vec{BA}|=\sqrt{4+4+9}   = √17
|\vec{BC}|=\sqrt{1+1+4}   = √6
Therefore,
cos θ = = \frac{\vec{BA}.\vec{BC}}{|\vec{BA}||\vec{BC}|}
cos θ = 10/ √(17×6)
θ = cos^-1(10/√102)

Question 25. If A, B, C have position vectors (0, 1, 1), (3, 1, 5), (0, 3, 3) respectively, show that △ABC is right-angled at C.

Solution:

Given, position vectors A(0, 1, 1), B(3, 1, 5), C(0, 3, 3)
Now, \vec{AC} = (0-0)\hat{i}+(3-1)\hat{j}+(3-1)\hat{k}
=2\hat{j}+2\hat{k}
\vec{BC}=(0-3)\hat{i}+(3-1)\hat{j}+(3-5)\hat{k}
= -3\hat{i}+2\hat{j}-2\hat{k}
\vec{AC}.\vec{BC}=(2\hat{j}+2\hat{k}).(-3\hat{i}+2\hat{j}-2\hat{k})
= 2 × 2 - 2 × 2 = 0
Thus, \vec{AC} and \vec{BC} are perpendicular hence △ABC is right-angled at C

Question 26. Find the projection of \vec{b}+\vec{c} on \vec{a} , where \vec{a}=2\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}+2\hat{j}-2\hat{k} and \vec{c}=2\hat{i}-\hat{j}+4\hat{k} .

Solution:

Given:
\vec{a}=2\hat{i}-2\hat{j}+\hat{k}
|\vec{a}|=\sqrt{4+4+1} = 3
\vec{b}=\hat{i}+2\hat{j}-2\hat{k}
\vec{c}=2\hat{i}-\hat{j}+4\hat{k}
To find the projection of \vec{b}+\vec{c}    on \vec{a}
\vec{b}+\vec{c} = 3\hat{i}+\hat{j}+2\hat{k}
Now, Projection of \vec{b}+\vec{c}   = [\frac{(\vec{b}+\vec{c})\vec{a}}{|\vec{a}|^2}]\vec{a}
= [\frac{6-2+2}{3^2}](2\hat{i}-2\hat{j}+\hat{k})
= 6/9 × 3
= 2

Question 27. If \vec{a}=5\hat{i}-\hat{j}-3\hat{k} and \vec{b}=\hat{i}+3\hat{j}-5\hat{k} , then show that the vectors \vec{a}+\vec{b} and \vec{a}-\vec{b} are orthogonal.

Solution:

Given: \vec{a}=5\hat{i}-\hat{j}-3\hat{k}
\vec{b}=\hat{i}+3\hat{j}-5\hat{k}
To prove
(\vec{a}+\vec{b})(\vec{a}-\vec{b})=0
Taking LHS
= |\vec{a}|^2-|\vec{b}|^2
= \sqrt{5^2+1+3^2}-\sqrt{1+3^2+5^2}
= √35 - √35
= 0
Thus, the given vectors \vec{a}+\vec{b} and \vec{a}-\vec{b} are orthogonal.

Question 28. A unit vector \vec{a} makes angle π/2 and π/3 with \hat{i} and \hat{j} respectively and an acute angle θ with \hat{k} . Find the angle θ and components of \vec{a} .

Solution:

Let us assume \vec{a}=a_1i + a_2j+a_3k
We know that
a₁²+ a₂²+ a₃² = 1 ....(1)
So,
\vec{a}i=a_1
|\vec{a}||i|cos\frac{π}{4}=a_1
(1)(1)(1/√2) = a₁
a₁= 1/√2
Again we take
\vec{a}j=a_2
|\vec{a}||j|cos\frac{π}{3}=a_2
(1)(1)(1/2) = a₂
a₂= 1/2
Put all these values in eq(1) to find the value of a₃
(1/√2)²+ (1/2)²+ a₃² = 1 ....(1)
a₃² = 1/4
a₃ = 1/2
Now we find the value of θ
\vec{a}k=a_3
|\vec{a}||k|cos\theta=a_3
(1)(1)cosθ = 1/2
cosθ = 1/2
cosθ = π/3
and components of \vec{a}
\vec{a}=\frac{1}{\sqrt{2}}i + \frac{1}{2}j+\frac{1}{2}k

Question 29. If two vectors \vec{a} and \vec{b} are such that |\vec{a}| = 2, |\vec{b}| = 1, and \vec{a}.\vec{b} =1. Find the value of (3\vec{a}-5\vec{b}).(2\vec{a}+7\vec{b}).

Solution:

Given, (3\vec{a}-5\vec{b}).(2\vec{a}+7\vec{b})
=6|\vec{a}|^2+21\vec{a}.\vec{b}-10\vec{a}.\vec{b}-35|\vec{b}|^2
=6|\vec{a}|^2+11\vec{a}.\vec{b}-35|\vec{b}|^2
= 6(2)² + 11(1) - 35(1)²
= 24 + 11 - 35
= 35 - 35 = 0

Question 30. If \vec{a} is a unit vector, then find |\vec{x}| in each of the following:

(i) (\vec{x}-\vec{a})(\vec{x}+\vec{a})=8

Solution:

Given, (\vec{x}-\vec{a})(\vec{x}+\vec{a})=8
⇒|\vec{x}|^2-\vec{x}.\vec{a}+\vec{a}\vec{x}+|\vec{a}|^2=8
⇒ |\vec{x}|^2-|\vec{a}|^2=8
⇒ |\vec{x}|^2-1=8
⇒ |\vec{x}|^2=9
⇒|\vec{x}|=3

(ii) (\vec{x}-\vec{a})(\vec{x}+\vec{a})=12

Solution:

Given, (\vec{x}-\vec{a})(\vec{x}+\vec{a})=12
⇒ |\vec{x}|^2-\vec{x}.\vec{a}+\vec{a}\vec{x}+|\vec{a}|^2=12
⇒ |\vec{x}|^2-|\vec{a}|^2=12
⇒ |\vec{x}|^2-1=12
⇒ |\vec{x}|^2=13
⇒ |\vec{x}| =√13

Question 31. Find |\vec{a}| and |\vec{b}| , if

(i) (\vec{a}+\vec{b})(\vec{a}-\vec{b}) = 12 and |\vec{a}| = 2 |\vec{b}|

Solution:

Given, (\vec{a}+\vec{b})(\vec{a}-\vec{b})   = 12
⇒  |\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=12
⇒  |\vec{a}|^2-|\vec{b}|^2    = 12
⇒ 4|\vec{b}|^2-|\vec{b}|^2    = 12
⇒ 3|\vec{b}|^2   = 12
⇒ |\vec{b}|   = 2
So,
|\vec{a}|   = 4

(ii) (\vec{a}+\vec{b})(\vec{a}-\vec{b}) = 8 and |\vec{a}| = 8|\vec{b}|

Solution:

Given, (\vec{a}+\vec{b})(\vec{a}-\vec{b})   = 8
⇒ |\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=8
⇒ |\vec{a}|^2-|\vec{b}|^2=8
⇒ 64|\vec{b}|^2-|\vec{b}|^2=8
⇒ 63|\vec{b}|^2=8
⇒ |\vec{b}|   = √(8/63)
So,
|\vec{a}|   = 8√(8/63)

(iii) (\vec{a}+\vec{b})(\vec{a}-\vec{b}) = 3 and |\vec{a}| = 2|\vec{b}|

Solution:

Given, (\vec{a}+\vec{b})(\vec{a}-\vec{b})= 3
⇒ |\vec{a}|^2-\vec{a}.\vec{b}+\vec{a}\vec{b}-|\vec{b}|^2=3
⇒ |\vec{a}|^2-|\vec{b}|^2 = 3
⇒ 4|\vec{b}|^2-|\vec{b}|^2=3
⇒ 3|\vec{b}|^2   = 3
⇒ |\vec{b}|   = 1
So,
|\vec{a}|   = 2

Question 32. Find |\vec{a}-\vec{b}| , if

(i) |\vec{a}| = 2, |\vec{b}| = 5 and \vec{a}.\vec{b} = 8

Solution:

We have, |\vec{a}-\vec{b}|
⇒ |\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2
⇒ |\vec{a}-\vec{b}|^2   = 2²- 2 × 8 + 5²
⇒ |\vec{a}-\vec{b}|^2   = 4 - 16 + 25
⇒ |\vec{a}-\vec{b}|^2   = 13
⇒ |\vec{a}-\vec{b}|   = √13

(ii) |\vec{a}| = 3, |\vec{b}| = 4 and \vec{a}.\vec{b} = 1

Solution:

We have, |\vec{a}-\vec{b}|
⇒  |\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2
⇒ |\vec{a}-\vec{b}|^2    = 3²- 2 × 1 + 4²
⇒ |\vec{a}-\vec{b}|^2    = 9 - 2 + 16
⇒ |\vec{a}-\vec{b}|^2   = 23
⇒ |\vec{a}-\vec{b}|    = √23

(iii) |\vec{a}|=2, |\vec{b}| = 3 and \vec{a}.\vec{b} = 4

Solution:

We have,  |\vec{a}-\vec{b}|
⇒ |\vec{a}-\vec{b}|^2 =|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2
⇒ |\vec{a}-\vec{b}|^2   = 2²- 2 × 4 + 3²
⇒ |\vec{a}-\vec{b}|^2   = 4 - 8 + 9
⇒ |\vec{a}-\vec{b}|^2   = 5
⇒ |\vec{a}-\vec{b}|    = √5

Practice Problems on Scalar or Dot Product

1. Find the scalar product of vectors a = 2i + 3j - k and b = i - 2j + 4k.

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

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

4. Find the angle between vectors a = 2i + 2j + k and b = i + j + 2k.

5. Calculate the projection of vector a = 3i - 4j + 2k onto vector b = i + 2j - 2k.

6. If a · b = 10 and |a| = 3, |b| = 5, find the angle between a and b.

7. Prove that (a + b) · (a - b) = |a|² - |b|².

8. Find a unit vector perpendicular to both a = i + j + k and b = 2i - j + 2k.

9. If a = 2i - j + 3k and b = i + 2j - k, find a scalar λ such that (a + λb) · b = 0.

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

Summary

Chapter 24 of RD Sharma Class 12 Solutions focuses on the Scalar or Dot Product of vectors. Key concepts covered include:

Definition of scalar product
Properties of scalar product
Geometric interpretation of scalar product
Applications in finding angles between vectors
Projections of vectors
Conditions for perpendicularity and parallelism of vectors
Vector equations using scalar product

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

Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.1 | Set 2

Question 17. If \vec{α}=3\hat{i}+4\hat{j}+5\hat{k} and \vec{β}=2\hat{i}+\hat{j}-4\hat{k} , then express \vec{β} in the form of \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} .

Question 18. If either \vec{a}=\vec{0} or \vec{b}=\vec{0} , then \vec{a}.\vec{b}=0 . But, The converse need not be true. Justify your answer with an example.

Question 19. Show that the vectors \vec{a}=3\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}-3\hat{j}+5\hat{k},\vec{c}=2\hat{i}+\hat{j}-4\hat{k} form a right-angled triangle.

Question 20. If \vec{a}=2\hat{i}+2\hat{j}+3\hat{k} , \vec{b}=-\hat{i}+2\hat{j}+\hat{k} and \vec{c}=3\hat{i}+\hat{j} are such that \vec{a}+λ\vec{b} is perpendicular to \vec{c} , then find the value of λ.

Question 21. Find the angles of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4), and C(5, 7, 1).

Question 22. Find the magnitude of two vectors \vec{a} and \vec{b} , having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.

Question 23. Show that the points whose position vector are \vec{a} =4\hat{i}-3\hat{j}+\hat{k},\vec{b} =2\hat{i}-4\hat{j}+5\hat{k}, \vec{c} =\hat{i}-\hat{j} form a right triangle.

Question 24. If the vertices A, B, C of △ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, what is the magnitude of ∠ABC?

Question 25. If A, B, C have position vectors (0, 1, 1), (3, 1, 5), (0, 3, 3) respectively, show that △ABC is right-angled at C.

Question 26. Find the projection of \vec{b}+\vec{c} on \vec{a} , where \vec{a}=2\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}+2\hat{j}-2\hat{k} and \vec{c}=2\hat{i}-\hat{j}+4\hat{k} .

Question 27. If \vec{a}=5\hat{i}-\hat{j}-3\hat{k} and \vec{b}=\hat{i}+3\hat{j}-5\hat{k} , then show that the vectors \vec{a}+\vec{b} and \vec{a}-\vec{b} are orthogonal.

Question 28. A unit vector \vec{a} makes angle π/2 and π/3 with \hat{i} and \hat{j} respectively and an acute angle θ with \hat{k} . Find the angle θ and components of \vec{a} .

Question 29. If two vectors \vec{a} and \vec{b} are such that |\vec{a}| = 2, |\vec{b}| = 1, and \vec{a}.\vec{b} =1. Find the value of (3\vec{a}-5\vec{b}).(2\vec{a}+7\vec{b}).

Question 30. If \vec{a} is a unit vector, then find |\vec{x}| in each of the following:

(i) (\vec{x}-\vec{a})(\vec{x}+\vec{a})=8

(ii) (\vec{x}-\vec{a})(\vec{x}+\vec{a})=12

Question 31. Find |\vec{a}| and |\vec{b}| , if

(i) (\vec{a}+\vec{b})(\vec{a}-\vec{b}) = 12 and |\vec{a}| = 2 |\vec{b}|

(ii) (\vec{a}+\vec{b})(\vec{a}-\vec{b}) = 8 and |\vec{a}| = 8|\vec{b}|

(iii) (\vec{a}+\vec{b})(\vec{a}-\vec{b}) = 3 and |\vec{a}| = 2|\vec{b}|

Question 32. Find |\vec{a}-\vec{b}| , if

(i) |\vec{a}| = 2, |\vec{b}| = 5 and \vec{a}.\vec{b} = 8

(ii) |\vec{a}| = 3, |\vec{b}| = 4 and \vec{a}.\vec{b} = 1

(iii) |\vec{a}|=2, |\vec{b}| = 3 and \vec{a}.\vec{b} = 4

Practice Problems on Scalar or Dot Product

Summary

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