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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 28 |

Exercise | 28.8 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 23.8 Solutions Algebra of Vectors**

**Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed **

### Question 1. Show that the points whose position vectors are given are collinear:

### (i)

**Solution:**

Let x =

y =

z =

Then

= Position vector of (y) – Position vector of (x)

=

=

= Position vector of (z) – Position vector of (y)

=

=

As,

So, and are parallel vectors but y is a common point to them. Hence, the given points x, y, z are collinear.

### (ii)

**Solution:**

Let

x =

y =

z =

Then,

= Position vector of (y) – Position vector of (x)

=

=

= Position vector of (z) – Position vector of (y)

=

=

As,

So, and are parallel vectors but y is a common point to them. Hence, the given points x, y, z are collinear.

### Question 2 (i). Using vector method, prove that A(6, -7, -1), B(2, -3, 1), and C(4, -5, 0) are collinear.

**Solution:**

The points given are A(6, -7, -1), B(2, -3, 1), and C(4, -5, 0)

So,

= Position vector of (B) – Position vector of (A)

=

=

= Position vector of (C) – Position vector of (B)

=

=

As,

So, and are parallel vectors but B is a common point to them. Hence, the given points A, B, C are collinear.

### Question 2 (ii). Using vector method, prove that A(2, -1, 3), B(4, 3, 1), and C(3, 1, 2) are collinear.

**Solution:**

The points given are A(2, -1, 3), B(4, 3, 1), C(3, 1, 2)

So, the

= Position vector of (B) – Position vector of (A)

=

=

= Position vector of (C) – Position vector of (B)

=

=

As,

So, and are parallel vectors but B is a common point to them. Hence, the given points A, B, C are collinear.

### Question 2 (iii). Using vector method, prove that X(1, 2, 7), Y(2, 6, 3), and Z(3, 10, -1) are collinear.

**Solution:**

The points given are X(1, 2, 7), Y(2, 6, 3), Z(3, 10, -1).

So, the

= Position vector of (Y) – Position vector of (X)

=

=

= = Position vector of (Z) – Position vector of (Y)

=

=

As,

So, and are parallel vectors but Y is a common point to them. Hence, the given points X, Y, Z are collinear.

### Question 2 (iv). Using vector method, prove that X(-3, -2, -5), Y(1, 2, 3), and Z(3, 4, 7) are collinear.

**Solution:**

The given points are X(-3, -2, -5), Y(1, 2, 3), and Z(3, 4, 7)

So,

= Position vector of (Y) – Position vector of (X)

=

=

= = Position vector of (Z) – Position vector of (Y)

=

=

As,

So, and are parallel vectors but Y is a common point to them. Hence, the given points X, Y, Z are collinear.

### Question 2 (v). Using vector method, prove that X(2, -1, 3), Y(3, -5, 1), and Z(-1, 11, 9) are collinear.

**Solution:**

= Position vector of (Y) – Position vector of (X)

=

=

= = Position vector of (Z) – Position vector of (Y)

=

=

As,

So, and are parallel vectors but Y is a common point to them. Hence, the given points X, Y, Z are collinear.

### Question 3 (i). If are non-zero, non-coplaner vectors, prove that the vectors are coplanar.

**Solution:**

The given vectors are

X =

Y =

Z =

Three vectors are coplanar, if they satisfy the given conditions(for real u and v)

X = u * Y + v * Z

On comparing coefficients, we get the following equations

7u + 3v = 5 -(1)

20v – 8u = 6 -(2)

9u + 5v = 7 -(3)

From first two equations, we find that

u = 1/2

v = 1/2

Now put the value of u and v in eq(3)

9(1/2) + 5(1/2) = 7

14/2 = 7

7 = 7

So, the value satisfies the third equation.

Hence, the given vectors X, Y, Z are coplanar.

### Question 3 (ii). If are non-zero, non-coplaner vectors, prove that the vectors are coplanar.

**Solution:**

The given vectors are

X =

Y =

Z =

Three vectors are coplanar, if they satisfy the given conditions(for real u and v)

X = u * Y + v * Z

On comparing coefficients, we get the following equations

-2v = 1 -(1)

3v – 3u = -2 -(2)

5u – 4v = 3 -(3)

From the first two equations, we find that

v = -1/2

u = 1/6

Now put the value of u and v in eq(3)

5(1/6) – 4(-1/2) = 3

5/6 + 2 = 3

(5 + 12)/6 = 3

17/6 ≠ 3

The value doesn’t satisfy the third equation. Hence, the given vectors X, Y, Z are not coplanar.

### Question 4.Show that the four points having position vectors are coplanar.

**Solution:**

Let the given vectors be

= Position vector of (X) – Position vector of (W)

=

=

= Position vector of (Y) – Position vector of (W)

=

=

= Position vector of (Z) – Position vector of (W)

=

=

The given vectors are coplanar if,

WX = u(WY) + v(WZ)

On comparing coefficients, we get the following equations

-6u – 4v = 10 -(1)

10u + 2v = -12 -(2)

-6u + 10v = -4 -(3)

From the first two equations, we find that

u = -1

v = -1

Now put the value of u and v in eq(3)

-6(-1) + 10(-1) = -4

6 – 10 = -4

-4 = -4

The value satisfies the third equation. Hence, the given vectors W, X, Y, Z are coplanar.

### Question 5(i). Prove that the following vectors are coplanar Show that the points

**Solution:**

The given vectors are

The given vectors are coplanar if,

A = u(B) + v(C)

On comparing coefficients, we get the following equations

u + 3v = 2 -(1)

-3u – 4v = -1 -(2)

-5u – 4v = 1 -(3)

From the first two equations, we find that

u = -1

v = 1

Now put the value of u and v in eq(3)

-5(-1) – 4(1) = 1

5 – 4 = 1

1 = 1

The value satisfies the third equation. Hence, the given vectors A, B, C are coplanar.

### Question 5(ii). Prove that the following vectors are coplanar Show that the points

**Solution:**

The given vectors are

The given vectors are coplanar if,

A = u(B) + v(C)

On comparing coefficients, we get the following equations

2u – v = 1 -(1)

3u – 2v = 1 -(2)

-u + 2v = 1 -(3)

From the first two equations, we find that

u = 1

v = 1

Now put the value of u and v in eq(3)

-(1) + 2(1) = 1

1 = 1

The value satisfies the third equation. Hence, the given vectors A, B, C are coplanar.

### Question 6 (i). Prove that the vector are non-coplanar.

**Solution:**

The given vectors are

The given vectors are coplanar if,

A = u(B) + v(C)

On comparing coefficients, we get the following equations

2u + 7v = 3 -(1)

-u – v = 1 -(2)

7u + 23v = -1 -(3)

From the first two equations, we find that

u = -2

v = 1

Now put the value of u and v in eq(3)

7(-2) + 23(1) = -1

-14 + 23 = -1

-9 ≠ -1

The value does not satisfy the third equation. Hence, the given vectors A, B, C are not coplanar.

### Question 6 (ii). Prove that the vector are non-coplanar.

**Solution:**

The given vectors are

The given vectors are coplanar if,

A = u(B) + v(C)

On comparing coefficients, we get the following equations

2u + v = 1 -(1)

u + v = 2 -(2)

3u + v = 3 -(3)

From the first two equations, we find that

u = 0

v = 1

Now put the value of u and v in eq(3)

3(0) + 1 = 3

1 = 3

The value does not satisfy the third equation. Hence, the given vectors A, B, C are not coplanar.

### Question 7(i). If are non-coplanar vectors, prove that the given vectors are non-coplanar

**Solution:**

The given vectors are )

The given vectors are coplanar if,

D = u(E) + v(F)

On comparing coefficients, we get the following equations

u + v = 2 -(1)

u + v = -1 -(2)

-2u – 3v = 3 -(3)

There is no value that satisfies the third equation. Hence, the given vectors D, E, F are not coplanar.

### Question 7(ii). If are non-coplanar vectors, prove that the given vectors are non-coplanar

**Solution:**

The given vectors are

The given vectors are coplanar if,

D = u(E) + v(F)

On comparing coefficients, we get the following equations

2u + v = 1 -(1)

u + v = 2 -(2)

3u + v = 3 -(3)

From the first two equations, we find that

u = -1

v = 3

Now put the value of u and v in eq(3)

3(-1) + (3) = 3

0 = 3

There is no value that satisfies the third equation. Hence, the given vectors D, E, F are not coplanar.

### Question 8. Show that the vector given by are non-coplanar. Express vector \vec{d} = as a linear combination of the vector .

**Solution:**

The given vectors are

The given vectors are coplanar if,

D = u(E) + v(F)

On comparing coefficients, we get the following equations

2u + v = 1 -(1)

u + v = 2 -(2)

3u + v = 3 -(3)

From above two equations,

u = -1

v = 3

Now put the value of u and v in eq(3)

3(-1) + (3) = 3

0 = -3

There is no value that satisfies the third equation. Hence, the given vectors D, E, F are not coplanar

The given vectors are

The given vectors are coplanar if,

On comparing coefficients, we get the following equations,

x + 2y + z = 2 -(1)

2x + y + z = -1 -(2)

3x + 3y + z = -3 -(3)

From above three equations,

x = -8/3

y = 1/3

z = 4

Therefore,

### Question 9. Prove that a necessary and sufficient condition for the three vectors to be coplanar is that these exist scalar l, m, n, not all zero simultaneously such that

**Solution:**

Given conditions: Let us considered be three coplanar vectors.

Then one of them is expressible as a linear combination of other two vectors.

Let,

Here, l = x, y = m, n = -1

From above,

Hence, is a linear combination of two vectors .

Hence proved that are coplanar vectors.

### Question 10. Show that the four points A, B, C, and D with position vectors respectively are coplanar if and only if .

**Solution:**

Given: A, B, C, D be four vectors with position vector

Let us considered A, B, C, D be coplanar.

Then, there exists x, y, z, u not all zero such that,

Let us considered x = 3, y = -2, z = 1, y = -2

So,

and x + y + z + u = 3 – 2 + 1 – 2 = 0

So, A, B, C, D are coplanar.

Let us considered

Now on dividing both side by sum of coefficient 4

It shows that point P divides AC in the ratio 1:3 and BD in the ratio 2:2 internally,

hence P is the point of intersection of AC and BD.

So, A, B, C, D are coplanar.

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