Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 3

Last Updated : 28 Mar, 2021

(i) =âˆš3, = 2 and = âˆš6

Solution:

We know,

â‡’ âˆš6 = 2âˆš3 cos Î¸

â‡’ cos Î¸ = 1/âˆš2

â‡’ Î¸ = cos-1(1/âˆš2)

â‡’ Î¸ = Ï€/4

(ii) = 3, = 3 and  = 1

Solution:

We know,

â‡’ 1 = 3Ã—3 cos Î¸

â‡’ cos Î¸ = 1/9

â‡’ Î¸ = cos-1(1/9)

Question 34. Express the vector as the sum of two vectors such that one is parallel to the vector and other is perpendicular to

Solution:

Given,

Let the two vectors be

Now,    ….(1)

Assuming  is parallel to

Then,      ……(2)

is perpendicular to

Then,     ……(3)

From eq(1)

â‡’

â‡’

â‡’

From eq(3)

â‡’

â‡’ (5-3Î»)3+(5-Î»)=0

â‡’ 15-9Î»+5-Î»=0

â‡’ -10Î» = -20

â‡’ Î»=2

From eq(2)

Question 35. If and are two vectors of the same magnitude inclined at an angle of 30Â° such that = 3, find

Solution:

Given that two vectors of the same magnitude inclined at an angle of 30Â°, and

To find

We know,

â‡’ 3 =

â‡’ 3 =

â‡’ 3 = (âˆš3/2)

â‡’= 6/âˆš3

â‡’

Question 36. Express as the sum of a vector parallel and a vector perpendicular to

Solution:

Assuming

Let the two vectors be

Now,

or   ….(1)

Assumingis parallel to

then,       â€¦(2)

is perpendicular to

then,……(3)

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

â‡’

â‡’

â‡’

From eq(3)

â‡’

â‡’ (2 – 2Î»)2 – (1 + 4Î»)4 – (3 + 2Î»)2 = 0

â‡’ 4 – 4Î» – 4 – 16Î» – 6 – 4Î» = 0

â‡’ 24Î» = -6

â‡’ Î» = -6/24

From eq(2)

Question 37. Decompose the vector into vectors which are parallel and perpendicular to the vector

Solution:

Let and

Let be a vector parallel to

Therefore,

to be decomposed into two vectors

â‡’

â‡’

Now, is perpendicular to

or

â‡’

â‡’ 6 – Î» – 3 – Î» – 6 – Î» = 0

â‡’ Î» = -1

Therefore, the required vectors are  and

Question 38. Let and . Find Î» such that  is orthogonal to

Solution:

Given,

According to question

â‡’

â‡’

â‡’

â‡’ 25 + 1 + 49 = 1 + 1 + Î»2

â‡’ Î»2 = 73

â‡’ Î» = âˆš73

Question 39. If  and , what can you conclude about the vector ?

Solution:

Given, ,

Now,

We conclude that or  or Î¸ = 90Â°

Thus,  can be any arbitrary vector.

Question 40. If  is perpendicular to both  and , then prove that it is perpendicular to both  and

Solution:

Given  is perpendicular to both  and

….(1)

….(2)

To prove  and

Now,

â‡’       [From eq(1) and (2)]

Again,

â‡’     [From eq(1) and (2)]

Hence Proved

Solution:

Given,  and

To prove

Taking LHS

=

=

Taking RHS

=

=

LHS = RHS

Hence Proved

Question 42. If  are three non- coplanar vectors such that then show that  is the null vector.

Solution:

Given that

So either or

Similarly,

Either or

Also,

So or

But can’t be perpendicular to  and  because  are non-coplanar.

So  = 0 oris a null vector

Question 43. If a vector is perpendicular to two non- collinear vectors and , then is perpendicular to every vector in the plane of and

Solution:

Given that  is perpendicular to  and

Let be any vector in the plane of  and  and is the linear combination of  and

[x, y are scalars]

Now

â‡’

â‡’

â‡’

â‡’

Therefore, is perpendicular to  i.e.  is perpendicular to every vector.

Solution:

Given that

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’ cos Î¸ =

Question 45. Let  and be vector such .  = 3,  = 4 and  = 5, then find

Solution:

Given that and are vectors such that  = 3,  = 4 and =5,

To find

Taking

Squaring on both side, we get

â‡’

â‡’

â‡’

â‡’

â‡’

Therefore,

Question 46. Let and be three vectors. Find the values of x for which the angle between and is acute and the angle between and is obtuse.

Solution:

Given

Case I: When angle between and is acute:-

>0

â‡’

â‡’ x2 – 2 – 2 > 0

â‡’ x2 > 4

x âˆˆ (2, -2)

Case II: When angle between and is obtuse:-

â‡’

â‡’ x2 – 5 – 4 < 0

â‡’ x2 < 9

x âˆˆ (3, -3)

Therefore, x âˆˆ (-3, -2)âˆª(2, 3)

Question 47. Find the value of x and y if the vectorsand  are mutually perpendicular vectors of equal magnitude.

Solution:

Given are mutually perpendicular vectors of equal magnitude.

â‡’ 32 + x2 + (-1)2 = 22 + 12 + y2

â‡’ x2+10 = y2+5

â‡’ x2 – y2 + 5 = 0    ….(1)

Now,

â‡’ 6 + x – y = 0

â‡’ y = x + 6      …..(2)

From eq(1)

x2 – (x + 6)2 + 5 = 0

â‡’ x2 – (x2 + 36 – 12x) + 5 = 0

â‡’ -12x – 31 = 0

â‡’ x = -31/12

Now, y = -31/12 + 6

y = 41/12

Question 48. If and are two non-coplanar unit vectors such that , find

Solution:

Given that and are two non-coplanar unit vectors such that

To find

Now,

Now,

= 6 – 13(1/2) – 5

= 1 – 13/2

= -11/2

Question 49. If  are two vectors such that || = , then prove that  is perpendicular to

Solution:

To prove

Now,

Squaring on both side, we get

â‡’

â‡’

â‡’

â‡’

Therefore, is perpendicular to

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