Class 12 RD Sharma Solutions Chapter 24 Scalar or Dot Product Exercise 24.1 | Set 3 (original) (raw)

Last Updated : 27 Aug, 2024

Question 33. Find the angle between the two vectors \vec{a} and \vec{b} , if

(i) |\vec{a}| =√3, |\vec{b}| = 2 and \vec{a}.\vec{b} = √6

**Solution:

We know, \vec{a}\vec{b} = |\vec{a}||\vec{b}|cos\theta

⇒ √6 = 2√3 cos θ

⇒ cos θ = 1/√2

⇒ θ = cos-1(1/√2)

⇒ θ = π/4

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

**Solution:

We know, \vec{a}\vec{b} = |\vec{a}||\vec{b}|cos\theta

⇒ 1 = 3×3 cos θ

⇒ cos θ = 1/9

⇒ θ = cos-1(1/9)

Question 34. Express the vector \vec{a}=5\hat{i}-2\hat{j}+5\hat{k} as the sum of two vectors such that one is parallel to the vector \vec{b}=3\hat{i}+\hat{k} and other is perpendicular to \vec{b}

**Solution:

Given, \vec{a}=5\hat{i}-2\hat{j}+5\hat{k}

\vec{b}=3\hat{i}+\hat{k}

Let the two vectors be \vec{a_1}, \vec{a_2}

Now, \vec{a}=\vec{a_1}+\vec{a_2} ....(1)

Assuming \vec{a_1} is parallel to \vec{b}

Then, \vec{a_1}=λ\vec{b} ......(2)

\vec{a_2} is perpendicular to \vec{b}

Then, \vec{a_2}.\vec{b}=0 ......(3)

From eq(1)

\vec{a_1}=λ\vec{b}

⇒ \vec{a_2} = \vec{a}-λ\vec{b}

⇒ \vec{a_2} =5\hat{i}-2\hat{j}+5\hat{k} -λ(3\hat{i}+\hat{k})

⇒ \vec{a_2} =(5-3λ)\hat{i}-2\hat{j}+(5-λ)\hat{k}

From eq(3)

\vec{a_2}.\vec{b}=0

⇒ ((5-3λ)\hat{i}-2\hat{j}+(5-λ)\hat{k})(3\hat{i}+\hat{k})

⇒ (5-3λ)3+(5-λ)=0

⇒ 15-9λ+5-λ=0

⇒ -10λ = -20

⇒ λ=2

From eq(2)

\vec{a_1}=6\hat{i}+2\hat{k}

\vec{a_2}=-\hat{i}-2\hat{j}+3\hat{k}

\vec{a}=(6\hat{i}+2\hat{k})+(-\hat{i}-2\hat{j}+3\hat{k})

Question 35. If \vec{a} and \vec{b} are two vectors of the same magnitude inclined at an angle of 30° such that \vec{a}.\vec{b} = 3, find |\vec{a}|, |\vec{b}|.

**Solution:

Given that two vectors of the same magnitude inclined at an angle of 30°, and \vec{a}.\vec{b} = 3

To find |\vec{a}|, |\vec{b}|

We know, \vec{a}\vec{b} = |\vec{a}||\vec{b}|cos\theta

⇒ 3 = |\vec{a}||\vec{a}|cos θ

⇒ 3 = |\vec{a}|^2 cos 30°

⇒ 3 = |\vec{a}|^2 (√3/2)

⇒ |\vec{a}|^2 = 6/√3

⇒|\vec{a}|=|\vec{b}|=\sqrt{2\sqrt{3}}

Question 36. Express 2\hat{i}-\hat{j}+3\hat{k} as the sum of a vector parallel and a vector perpendicular to 2\hat{i}+4\hat{j}-2\hat{k}.

**Solution:

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

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

Let the two vectors be \vec{a_1}, \vec{a_2}

Now, \vec{a}=\vec{a_1}+\vec{a_2}

or \vec{a_2}=\vec{a}-\vec{a_1} ....(1)

Assuming \vec{a_1} is parallel to \vec{b}

then, \vec{a_1}=λ\vec{b} …(2)

\vec{a_2} is perpendicular to \vec{b}

then,\vec{a_2}.\vec{b}=0 ......(3)

Putting eq(2) in eq(1), we get

\vec{a_1}=λ\vec{b}

⇒ \vec{a_2} = \vec{a}-λ\vec{b}

⇒ \vec{a_2} =2\hat{i}-\hat{j}+3\hat{k} -λ(2\hat{i}+4\vec{j}-2\hat{k})

⇒ \vec{a_2} =(2-2λ)\hat{i}-(1+4λ)\hat{j}+(3+2λ)\hat{k}

From eq(3)

\vec{a_2}.\vec{b}=0

⇒((2-2λ)\hat{i}-(1+4λ)\hat{j}+(3+2λ)\hat{k})(2\hat{i}+4\hat{j}-2\hat{k})=0

⇒ (2 - 2λ)2 - (1 + 4λ)4 - (3 + 2λ)2 = 0

⇒ 4 - 4λ - 4 - 16λ - 6 - 4λ = 0

⇒ 24λ = -6

⇒ λ = -6/24

From eq(2)

\vec{a_1}=-1/4(2\hat{i}+4\hat{j}-2\hat{k})

\vec{a_1}=-1/2\hat{i}-1\hat{j}+1/2\hat{k}

\vec{a_2}=(2+1/2)\hat{i}-0\hat{j}+(3-1/2)\hat{k}

\vec{a_2}=5/2\hat{i}+5/2\hat{k}

\vec{a}=\vec{a_1}+\vec{a_2}

\vec{a}=(-1/2\hat{i}-1\hat{j}+1/2\hat{k})+(5/2\hat{i}+5/2\hat{k})

Question 37. Decompose the vector 6\hat{i}-3\hat{j}-6\hat{k} into vectors which are parallel and perpendicular to the vector \hat{i}+\hat{j}+\hat{k}.

**Solution:

Let \vec{a}=6\hat{i}-3\hat{j}-6\hat{k} and \vec{m}=\hat{i}+\hat{j}+\hat{k}

Let \vec{b} be a vector parallel to \hat{i}+\hat{j}+\hat{k}.

Therefore, \vec{b} =λ(\hat{i}+\hat{j}+\hat{k})

\vec{a} to be decomposed into two vectors

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

⇒ \vec{c}=\vec{a}-\vec{b}

⇒ \vec{c}=(6-λ)\hat{i}+(-3-λ)\hat{j}+(-6-λ)\vec{k}

Now, \vec{c} is perpendicular to \vec{m}

or \vec{c}.\vec{m}=0

⇒((6-λ)\hat{i}+(-3-λ)\hat{j}+(-6-λ)\vec{k})(\hat{i}+\hat{j}+\hat{k})=0

⇒ 6 - λ - 3 - λ - 6 - λ = 0

⇒ λ = -1

Therefore, the required vectors are \vec{b} =-\hat{i}-\hat{j}-\hat{k} and \vec{c}=7\hat{i}-2\hat{j}-5\vec{k}

Question 38. Let \vec{a}=5\hat{i}-\hat{j}+7\hat{k} and \vec{b}=\hat{i}-\hat{j}+λ\hat{k} . Find λ such that \vec{a}+\vec{b} is orthogonal to \vec{a}-\vec{b}

**Solution:

Given, \vec{a}=5\hat{i}-\hat{j}+7\hat{k}

\vec{b}=\hat{i}-\hat{j}+λ\hat{k}

According to question

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

⇒|\vec{a}|^2-|\vec{b}|^2=0

⇒ |\vec{a}|^2=|\vec{b}|^2

⇒ \sqrt{5^2+(-1)^2+7^2}=\sqrt{1^2+(-1)^2+λ^2}

⇒ 25 + 1 + 49 = 1 + 1 + λ2

⇒ λ2 = 73

⇒ λ = √73

Question 39. If \vec{a}.\vec{a}=0 and \vec{a}.\vec{b}=0 , what can you conclude about the vector \vec{b} ?

**Solution:

Given, \vec{a}.\vec{a}=0 ,\vec{a}.\vec{b}=0

|\vec{a}| = 0

Now, \vec{a}.\vec{b}=0

We conclude that \vec{a}=0 or \vec{b}=0 or θ = 90°

Thus, \vec{b} can be any arbitrary vector.

Question 40. If \vec{c} is perpendicular to both \vec{a} and \vec{b} , then prove that it is perpendicular to both \vec{a}+\vec{b} and \vec{a}-\vec{b}

**Solution:

Given \vec{c} is perpendicular to both \vec{a} and \vec{b}

\vec{c}.\vec{a}=0 ....(1)

\vec{c}.\vec{b}=0 ....(2)

To prove \vec{c}.(\vec{a}+\vec{b})=0 and \vec{c}.(\vec{a}-\vec{b})=0

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

⇒ \vec{c}.\vec{a}+\vec{c}.\vec{b}=0 [From eq(1) and (2)]

Again, \vec{c}.(\vec{a}-\vec{b})

⇒ \vec{c}.\vec{a}-\vec{c}.\vec{b}=0 [From eq(1) and (2)]

Hence Proved

Question 41. If |\vec{a}|= a and |\vec{b}|= b , prove that (\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2= (\frac{\vec{a}-\vec{b}}{ab})^2

**Solution:

Given, |\vec{a}| = a and |\vec{b}| = b,

To prove

(\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2= (\frac{\vec{a}-\vec{b}}{ab})^2.

Taking LHS

(\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2

=\frac{\vec{a}.\vec{a}}{a^4}+\frac{\vec{b}.\vec{b}}{b^4}-2\frac{\vec{a}.\vec{b}}{a^2b^2}

=\frac{a^2}{a^4}+\frac{b^2}{b^4}-2\frac{\vec{a}.\vec{b}}{a^2b^2}

= \frac{1}{a^2}+\frac{1}{b^2}-2\frac{\vec{a}.\vec{b}}{a^2b^2}

Taking RHS

(\frac{\vec{a}-\vec{b}}{ab})^2 \vec{d}.\vec{a}=0

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

=\frac{1}{a^2}+\frac{1}{b^2}-2\frac{\vec{a}.\vec{b}}{a^2b^2}

LHS = RHS

Hence Proved

Question 42. If \vec{a},\vec{b},\vec{c} are three non- coplanar vectors such that \vec{d}.\vec{a}=\vec{d}.\vec{b}=\vec{d}.\vec{c} =0 then show that \vec{d} is the null vector.

**Solution:

Given that \vec{d}.\vec{a}=0

So either \vec{d} = 0 or \vec{d}⊥\vec{a}=0

Similarly, \vec{d}.\vec{b} = 0

Either \vec{d}= 0 or \vec{d}⊥\vec{b}=0

Also, \vec{d}.\vec{c} =0

So \vec{d} = 0 or \vec{d}⊥\vec{c}=0

But \vec{d} can't be perpendicular to \vec{a},\vec{b} and \vec{c} because \vec{a},\vec{b},\vec{c} are non-coplanar.

So \vec{d} = 0 or \vec{d} is a null vector

Question 43. If a vector \vec{a} is perpendicular to two non- collinear vectors \vec{b} and \vec{c} , then is \vec{a} perpendicular to every vector in the plane of \vec{b} and \vec{c}

**Solution:

Given that \vec{a} is perpendicular to \vec{b} and \vec{c}

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

Let \vec{r} be any vector in the plane of \vec{b} and \vec{c} and \vec{r} is the linear combination of \vec{b} and \vec{c}

\vec{r} =x\vec{b}+y\vec{c} [x, y are scalars]

Now, \vec{a}.\vec{r}

⇒ \vec{a}.\vec{r} =\vec{a}(x\vec{b}+y\vec{c})

⇒ \vec{a}.\vec{r} =x(\vec{a}.\vec{b})+y(\vec{a}.\vec{c})

⇒ \vec{a}.\vec{r} =x0+y0

⇒ \vec{a}.\vec{r} = 0

Therefore, \vec{a} is perpendicular to \vec{r} i.e. \vec{a} is perpendicular to every vector.

Question 44. If \vec{a}+\vec{b}+\vec{c}=\vec{0} , how that the angle θ between the vectors \vec{b} and \vec{c} is given by cos θ = \frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}

**Solution:

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

⇒ \vec{a}=-(\vec{b}+\vec{c})

⇒ (\vec{a})^2=(\vec{b}+\vec{c})^2

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

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

⇒ |\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2=2|\vec{b}||\vec{c}|

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

⇒ cos θ = \frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}.

Question 45. Let \vec{u},\vec{v} and \vec{w} be vector such \vec{u}+\vec{v}+\vec{w}=\vec{0} . |\vec{u}| = 3, |\vec{v}| = 4 and |\vec{w}| = 5, then find \vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}

**Solution:

Given that \vec{u},\vec{v} and \vec{w} are vectors such that \vec{u}+\vec{v}+\vec{w}=\vec{0} . |\vec{u}| = 3, |\vec{v}| = 4 and |\vec{w}| =5,

To find \vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}

Taking

\vec{u}+\vec{v}+\vec{w}=\vec{0}

Squaring on both side, we get

⇒ (\vec{u}+\vec{v}+\vec{w})^2=\vec{0}

⇒ |\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2+2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=0

⇒ 2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=-(|\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2)

⇒ 2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=-(3^2+4^2+5^2)

⇒ 2(\vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u})=-50

Therefore, \vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}=-25

Question 46. Let \vec{a}=x^2\hat{i}+2\hat{j}-2\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}, and \vec{c}=x^2\hat{i}+5\hat{j}-4\hat{k} be three vectors. Find the values of x for which the angle between \vec{a} and \vec{b} is acute and the angle between \vec{b} and \vec{c} is obtuse.

**Solution:

Given \vec{a}=x^2\hat{i}+2\hat{j}-2\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}, \vec{c}=x^2\hat{i}+5\hat{j}-4\hat{k}

Case I: When angle between \vec{a} and \vec{b} is acute:-

\vec{a}.\vec{b} >0

⇒ (x^2\hat{i}+2\hat{j}-2\hat{k})(\hat{i}-\hat{j}+\hat{k})>0

⇒ x2 - 2 - 2 > 0

⇒ x2 > 4

x ∈ (2, -2)

Case II: When angle between \vec{b} and \vec{c} is obtuse:-

\vec{b}.\vec{c}<0

⇒ (\hat{i}-\hat{j}+\hat{k})(x^2\hat{i}+5\hat{j}-4\hat{k})<0

⇒ x2 - 5 - 4 < 0

⇒ x2 < 9

x ∈ (3, -3)

Therefore, x ∈ (-3, -2)∪(2, 3)

Question 47. Find the value of x and y if the vectors \vec{a}=3\hat{i}+x\hat{j}-\hat{k} and \vec{b}=2\hat{i}+\hat{j}+y\hat{k} are mutually perpendicular vectors of equal magnitude.

**Solution:

Given \vec{a}=3\hat{i}+x\hat{j}-\hat{k}, \vec{b}=2\hat{i}+\hat{j}+y\hat{k} are mutually perpendicular vectors of equal magnitude.

|\vec{a}|^2=|\vec{b}|^2

⇒ 32 + x2 + (-1)2 = 22 + 12 + y2

⇒ x2+10 = y2+5

⇒ x2 - y2 + 5 = 0 ....(1)

Now, \vec{a}.\vec{b} = 0

⇒ 6 + x - y = 0

⇒ y = x + 6 .....(2)

From eq(1)

x2 - (x + 6)2 + 5 = 0

⇒ x2 - (x2 + 36 - 12x) + 5 = 0

⇒ -12x - 31 = 0

⇒ x = -31/12

Now, y = -31/12 + 6

y = 41/12

Question 48. If \vec{a} and \vec{b} are two non-coplanar unit vectors such that |\vec{a}+\vec{b}| =√3 , find (2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}).

**Solution:

Given that \vec{a} and \vec{b} are two non-coplanar unit vectors such that |\vec{a}+\vec{b}| =√3

To find (2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}).

Now,

|\vec{a}+\vec{b}|^2 =3

⇒|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}=3

⇒ 1+1+2\vec{a}.\vec{b}=3

⇒ \vec{a}.\vec{b}=1/2

Now, (2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b})

= 6|\vec{a}|^2-13\vec{a}.\vec{b}-5|\vec{b}|^2

= 6 - 13(1/2) - 5

= 1 - 13/2

= -11/2

Question 49. If \vec{a},\vec{b} are two vectors such that |\vec{a}+\vec{b} | = |\vec{b}| , then prove that \vec{a}+2\vec{b} is perpendicular to \vec{a}.

**Solution:

To prove

(\vec{a}+2\vec{b})\vec{a}= 0

Now,

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

Squaring on both side, we get

|\vec{a}+\vec{b}|^2=|\vec{b}|^2

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

⇒ \vec{a}\vec{a}+\vec{a}\vec{b}+\vec{b}\vec{a}+\vec{b}\vec{b} = \vec{b}.\vec{b}

⇒ \vec{a}\vec{a}+2\vec{a}\vec{b} = 0

⇒ \vec{a}(\vec{a}+2\vec{b}) = 0

Therefore, \vec{a}+2\vec{b} is perpendicular to \vec{a}

Summary

This set focuses on more advanced applications of the dot product.

**Key points include:

Using dot product to solve complex geometric problems
Applying dot product properties in vector equations
Proving vector identities involving dot products
Utilizing dot product in problems related to direction cosines and ratios
Combining dot product with other vector operations

Practice Questions

**1). If a, b, c are unit vectors such that a + b + c = 0, prove that a · b = b · c = c · a = -1/2.

**2). Find the value of λ for which the vectors 2i + j - k, i - 2j + 3k, and λi + 2j - k are coplanar.

**3). Prove that (a × b) · (b × c) = (a · b)(b · c) - b²(a · c).

**4). If a · b = 3, b · c = 4, c · a = 5, |a| = 2, |b| = 3, and |c| = 4, find the value of |a × b + b × c + c × a|.

**5). Show that (a - b) · (a + b) = |a|² - |b|² for any two vectors a and b.

**6). If a, b, c are non-coplanar vectors and p = λa + μb + νc, express λ, μ, ν in terms of dot products.

**7). Prove that the sum of squares of the direction cosines of a line is always 1.

**8). If a = 3i - 2j + k, b = i + 2j - 2k, and c = 2i + j + 3k, find the angle between a + b and a - c.

**9). Show that (a · b)² + |a × b|² = |a|²|b|² for any two vectors a and b.

**10). If a, b, c are unit vectors such that a · b = b · c = c · a = 1/2, find the value of |a + b + c|.

Class 12 RD Sharma Solutions Chapter 24 Scalar or Dot Product Exercise 24.1 | Set 3 (original) (raw)

Question 33. Find the angle between the two vectors \vec{a} and \vec{b} , if

(i) |\vec{a}| =√3, |\vec{b}| = 2 and \vec{a}.\vec{b} = √6

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

Question 34. Express the vector \vec{a}=5\hat{i}-2\hat{j}+5\hat{k} as the sum of two vectors such that one is parallel to the vector \vec{b}=3\hat{i}+\hat{k} and other is perpendicular to \vec{b}

**Question 35. If \vec{a} **and \vec{b} **are two vectors of the same magnitude inclined at an angle of 30° such that \vec{a}.\vec{b} **= 3, find |\vec{a}|, |\vec{b}|.

Question 36. Express 2\hat{i}-\hat{j}+3\hat{k} as the sum of a vector parallel and a vector perpendicular to 2\hat{i}+4\hat{j}-2\hat{k}.

Question 37. Decompose the vector 6\hat{i}-3\hat{j}-6\hat{k} into vectors which are parallel and perpendicular to the vector \hat{i}+\hat{j}+\hat{k}.

Question 38. Let \vec{a}=5\hat{i}-\hat{j}+7\hat{k} and \vec{b}=\hat{i}-\hat{j}+λ\hat{k} . Find λ such that \vec{a}+\vec{b} is orthogonal to \vec{a}-\vec{b}

Question 39. If \vec{a}.\vec{a}=0 and \vec{a}.\vec{b}=0 , what can you conclude about the vector \vec{b} ?

Question 40. If \vec{c} is perpendicular to both \vec{a} and \vec{b} , then prove that it is perpendicular to both \vec{a}+\vec{b} and \vec{a}-\vec{b}

Question 41. If |\vec{a}|= a and |\vec{b}|= b , prove that (\frac{\vec{a}}{a^2}-\frac{\vec{b}}{b^2})^2= (\frac{\vec{a}-\vec{b}}{ab})^2

Question 42. If \vec{a},\vec{b},\vec{c} are three non- coplanar vectors such that \vec{d}.\vec{a}=\vec{d}.\vec{b}=\vec{d}.\vec{c} =0 then show that \vec{d} is the null vector.

Question 43. If a vector \vec{a} is perpendicular to two non- collinear vectors \vec{b} and \vec{c} , then is \vec{a} perpendicular to every vector in the plane of \vec{b} and \vec{c}

Question 44. If \vec{a}+\vec{b}+\vec{c}=\vec{0} , how that the angle θ between the vectors \vec{b} and \vec{c} is given by cos θ = \frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}

Question 45. Let \vec{u},\vec{v} and \vec{w} be vector such \vec{u}+\vec{v}+\vec{w}=\vec{0} . |\vec{u}| = 3, |\vec{v}| = 4 and |\vec{w}| = 5, then find \vec{u}.\vec{v}+\vec{v}.\vec{w}+\vec{w}.\vec{u}

Question 46. Let \vec{a}=x^2\hat{i}+2\hat{j}-2\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}, and \vec{c}=x^2\hat{i}+5\hat{j}-4\hat{k} be three vectors. Find the values of x for which the angle between \vec{a} and \vec{b} is acute and the angle between \vec{b} and \vec{c} is obtuse.

Question 47. Find the value of x and y if the vectors \vec{a}=3\hat{i}+x\hat{j}-\hat{k} and \vec{b}=2\hat{i}+\hat{j}+y\hat{k} are mutually perpendicular vectors of equal magnitude.

Question 48. If \vec{a} and \vec{b} are two non-coplanar unit vectors such that |\vec{a}+\vec{b}| =√3 , find (2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}).

Question 49. If \vec{a},\vec{b} are two vectors such that |\vec{a}+\vec{b} | = |\vec{b}| , then prove that \vec{a}+2\vec{b} is perpendicular to \vec{a}.

Summary

Practice Questions

Question 35. If \vec{a} and \vec{b} are two vectors of the same magnitude inclined at an angle of 30° such that \vec{a}.\vec{b} = 3, find |\vec{a}|, |\vec{b}|.