RD Sharma Class 12 Ex 23.8 Solutions Algebra of Vectors

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter28
Exercise28.8
CategoryRD 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) 2\hat{i}+\hat{j}-\hat{k}, 3\hat{i}-2\hat{j}+\hat{k}, and \ \hat{i}+4\hat{j}-3\hat{k}

Solution:

Let x = 2\hat{i}+\hat{j}-\hat{k}

y = 3\hat{i}-2\hat{j}+\hat{k}

z = \hat{i}+4\hat{j}-3\hat{k}

Then 

\overrightarrow{xy}  = Position vector of (y) – Position vector of (x)

3\hat{i}-2\hat{j}+\hat{k}-2\hat{i}-\hat{j}+\hat{k}

\hat{i}-3\hat{j}+2\hat{k}

\overrightarrow{yz}  = Position vector of (z) – Position vector of (y)

\hat{i}+4\hat{j}-3\hat{k} - 3\hat{i}+2\hat{j}-\hat{k}

-2\hat{i}+6\hat{j}-4\hat{k}

As, \overrightarrow{yz} = -2(\overrightarrow{xy})

So, \overrightarrow{yz}   and \overrightarrow{xy}  are parallel vectors but y is a common point to them. Hence, the given points x, y, z are collinear.

(ii) 3\hat{i}-2\hat{j}+4\hat{k}, \hat{i}+\hat{j}+\hat{k}, and \ -\hat{i}+4\hat{j}-2\hat{k}

Solution:

Let 

x = 3\hat{i}-2\hat{j}+4\hat{k}

y = \hat{i}+\hat{j}+\hat{k}

z = -\hat{i}+4\hat{j}-2\hat{k}

Then,

\overrightarrow{xy}  = Position vector of (y) – Position vector of (x)

\hat{i}+\hat{j}+\hat{k}-3\hat{i}+2\hat{j}-4\hat{k}

-2\hat{i}+3\hat{j}-3\hat{k}

\overrightarrow{yz}  = Position vector of (z) – Position vector of (y)

-\hat{i}+4\hat{j}-2\hat{k}-\hat{i}-\hat{j}-\hat{k}

-2\hat{i}+3\hat{j}-3\hat{k}

As, \overrightarrow{yz} = \overrightarrow{xy}

So, \overrightarrow{yz}   and \overrightarrow{xy}  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, \vec{A}=6\hat{i}-7\hat{j}-\hat{k}

\vec{B}=2\hat{i}-3\hat{j}+\hat{k}
\vec{C}=4\hat{i}-5\hat{j}

\overrightarrow{AB}  = Position vector of (B) – Position vector of (A)

2\hat{i}-3\hat{j}+\hat{k}-6\hat{i}+7\hat{j}+\hat{k}

-4\hat{i}+4\hat{j}+2\hat{k}

\overrightarrow{BC}  = Position vector of (C) – Position vector of (B)

=4\hat{i}-5\hat{j}-2\hat{i}+3\hat{j}-\hat{k}

2\hat{i}-2\hat{j}-\hat{k}

As, \overrightarrow{AB} = -2(\overrightarrow{BC})

So, \overrightarrow{AB}   and \overrightarrow{BC}  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 \vec{A}=2\hat{i}-\hat{j}+3\hat{k}

\vec{B}=4\hat{i}+3\hat{j}+\hat{k}
\vec{C}=3\hat{i}+\hat{j}+2\hat{k}

\overrightarrow{AB}   = Position vector of (B) – Position vector of (A)

4\hat{i}+3\hat{j}+\hat{k}-2\hat{i}+\hat{j}-3\hat{k}

2\hat{i}+4\hat{j}-2\hat{k}

\overrightarrow{BC}  = Position vector of (C) – Position vector of (B)

3\hat{i}+\hat{j}+2\hat{k}-4\hat{i}-3\hat{j}-\hat{k}

-\hat{i}-2\hat{j}+\hat{k}

As, \overrightarrow{AB} = -2(\overrightarrow{BC})

So, \overrightarrow{AB}   and \overrightarrow{BC}  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 \vec{X}=\hat{i}+2\hat{j}+7\hat{k}

\vec{Y}=2\hat{i}+6\hat{j}+3\hat{k}
\vec{Z}=3\hat{i}+10\hat{j}-\hat{k}

\overrightarrow{XY}  = Position vector of (Y) – Position vector of (X)

2\hat{i}+6\hat{j}+3\hat{k}-\hat{i}-2\hat{j}-7\hat{k}

\hat{i}+4\hat{j}-4\hat{k}

\overrightarrow{YZ}  = = Position vector of (Z) – Position vector of (Y)

3\hat{i}+10\hat{j}-\hat{k}-2\hat{i}-6\hat{j}-3\hat{k}

\hat{i}+4\hat{j}-4\hat{k}

As, \overrightarrow{YZ} = \overrightarrow{XY}

So, \overrightarrow{YZ}   and \overrightarrow{XY}  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, \vec{X}=-3\hat{i}-2\hat{j}-5\hat{k}

\vec{Y}=\hat{i}+2\hat{j}+3\hat{k}
\vec{Z}=3\hat{i}+4\hat{j}+7\hat{k}

\overrightarrow{XY}  = Position vector of (Y) – Position vector of (X)

\hat{i}+2\hat{j}+3\hat{k}+3\hat{i}+2\hat{j}+5\hat{k}

4\hat{i}+4\hat{j}+8\hat{k}

\overrightarrow{YZ}  = = Position vector of (Z) – Position vector of (Y)

3\hat{i}+4\hat{j}+7\hat{k}-\hat{i}-2\hat{j}-3\hat{k}

2\hat{i}+2\hat{j}+4\hat{k}

As, \overrightarrow{YZ} = 2(\overrightarrow{XY})

So, \overrightarrow{YZ}   and \overrightarrow{XY}  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:

\vec{X}=2\hat{i}-\hat{j}+3\hat{k}
\vec{Y}=3\hat{i}-5\hat{j}+\hat{k}
\vec{Z}=-\hat{i}+11\hat{j}+9\hat{k}

\overrightarrow{XY}  = Position vector of (Y) – Position vector of (X)

3\hat{i}-5\hat{j}+\hat{k}-2\hat{i}-\hat{j}+3\hat{k}

\hat{i}-4\hat{j}-2\hat{k}

\overrightarrow{YZ}  = = Position vector of (Z) – Position vector of (Y)

-\hat{i}+11\hat{j}+9\hat{k}-3\hat{i}+5\hat{j}-\hat{k}

-4\hat{i}+16\hat{j}+8\hat{k}

As, \overrightarrow{YZ} = 4(\overrightarrow{XY})

So, \overrightarrow{YZ}   and \overrightarrow{XY}  are parallel vectors but Y is a common point to them. Hence, the given points X, Y, Z are collinear.

Question 3 (i). If \vec{a},\vec{b},\vec{c}  are non-zero, non-coplaner vectors, prove that the vectors 5\vec{a}+6\vec{b}+7\vec{c}, 7\vec{a}-8\vec{b}+9\vec{c}, and \ 3\vec{a}+20\vec{b}+5\vec{c} are coplanar.

Solution:

The given vectors are 

X = 5\vec{a}+6\vec{b}+7\vec{c}

Y = 7\vec{a}-8\vec{b}+9\vec{c}

Z = 3\vec{a}+20\vec{b}+5\vec{c}

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

 X = u * Y + v * Z

5\vec{a}+6\vec{b}+7\vec{c}=u(7\vec{a}-8\vec{b}+9\vec{c})+v(3\vec{a}+20\vec{b}+5\vec{c})
5\vec{a}+6\vec{b}+7\vec{c}=\vec{a}(7u+3v)+\vec{b}(20v-8u)+\vec{c}(9u+5v)

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 \vec{a},\vec{b},\vec{c}  are non-zero, non-coplaner vectors, prove that the vectors \vec{a}-2\vec{b}+3\vec{c}, -3\vec{b}+5\vec{c}, and \ -2\vec{a}+3\vec{b}-4\vec{c}  are coplanar.

Solution:

The given vectors are 

X = \vec{a}-2\vec{b}+3\vec{c}

Y = -3\vec{b}+5\vec{c}

Z = -2\vec{a}+3\vec{b}-4\vec{c}

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

X = u * Y + v * Z

\vec{a}-2\vec{b}+3\vec{c}=u(-3\vec{b}+5\vec{c})+v(-2\vec{a}+3\vec{b}-4\vec{c})
\vec{a}-2\vec{b}+3\vec{c}=(-2v)\vec{a} +\vec{b}(-3u+3v)+\vec{c}(5u-4v)

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 6\hat{i}-7\hat{j}, 16\hat{i}-19\hat{j}-4\hat{k},3\hat{i}-6\hat{k},2\hat{i}-5\hat{j}+10\hat{k}  are coplanar.

Solution:

Let the given vectors be 

\vec{W}=6\hat{i}-7\hat{j}\\ \vec{X}=16\hat{i}-19\hat{j}-4\hat{k}\\ \vec{Y}=3\hat{i}-6\hat{k}\\ \vec{Z}=2\hat{i}-5\hat{j}+10\hat{k}

\overrightarrow{WX}  = Position vector of (X) – Position vector of (W)

16\hat{i}-19\hat{j}-4\hat{k} - 6\hat{i}+7\hat{j}

10\hat{i}-12\hat{j}-4\hat{k}

\overrightarrow{WY}  = Position vector of (Y) – Position vector of (W)

= 3\hat{i}-6\hat{k}- 6\hat{i}+7\hat{j}

-3\hat{i}+7\hat{j}-6\hat{k}

\overrightarrow{WZ}  = Position vector of (Z) – Position vector of (W)

2\hat{i}-5\hat{j}+10\hat{k} - 6\hat{i}+7\hat{j}

-4\hat{i}+2\hat{j}+10\hat{k}

The given vectors are coplanar if,

WX = u(WY) + v(WZ)

10\hat{i}-12\hat{j}-4\hat{k}=u(-6\hat{i}+10\hat{j}-6\hat{k})+v(-4\hat{i}+2\hat{j}+10\hat{k})
10\hat{i}-12\hat{j}-4\hat{k}=\hat{i}(-6u-4v)+\hat{j}(10u+2v)+\hat{k}(-6u+10v)

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 2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k},3\hat{i}-4\hat{j}-4\hat{k}

Solution:

The given vectors are \vec{A}=2\hat{i}-\hat{j}+\hat{k}\\ \vec{B}=\hat{i}-3\hat{j}-5\hat{k}\\ \vec{C}=3\hat{i}-4\hat{j}-4\hat{k}  

The given vectors are coplanar if,

A = u(B) + v(C)

2\hat{i}-\hat{j}+\hat{k}=u(\hat{i}-3\hat{j}-5\hat{k})+v(3\hat{i}-4\hat{j}-4\hat{k})
2\hat{i}-\hat{j}+\hat{k}=\hat{i}(u+3v)+\hat{j}(-3u-4v)+\hat{k}(-5u-4v)

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 \hat{i}+\hat{j}+\hat{k},2\hat{i}+3\hat{j}-\hat{k},-\hat{i}-2\hat{j}+2\hat{k}

Solution:

The given vectors are \vec{A}=\hat{i}+\hat{j}+\hat{k}\\ \vec{B}=2\hat{i}+3\hat{j}-\hat{k}\\ \vec{C}=-\hat{i}-2\hat{j}+2\hat{k}

The given vectors are coplanar if,

A = u(B) + v(C)

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

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 3\hat{i}+\hat{j}-\hat{k},2\hat{i}-\hat{j}+7\hat{k},7\hat{i}-\hat{j}+23\hat{k}  are non-coplanar.

Solution:

The given vectors are \vec{A}=3\hat{i}+\hat{j}-\hat{k}\\ \vec{B}=2\hat{i}-\hat{j}+7\hat{k}\\ \vec{C}=7\hat{i}-\hat{j}+23\hat{k}

The given vectors are coplanar if,

A = u(B) + v(C)

3\hat{i}+\hat{j}-\hat{k}=u(2\hat{i}-\hat{j}+7\hat{k})+v(7\hat{i}-\hat{j}+23\hat{k})
3\hat{i}+\hat{j}-\hat{k}=(2u+7v)\hat{i}+(-u-v)\hat{j}+(7u+23v)\hat{k}

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 \hat{i}+2\hat{j}+3\hat{k},2\hat{i}+\hat{j}+3\hat{k},\hat{i}+\hat{j}+\hat{k}  are non-coplanar.

Solution:

The given vectors are \vec{A}=\hat{i}+2\hat{j}+3\hat{k}\\ \vec{B}=2\hat{i}+\hat{j}+3\hat{k}\\ \vec{C}=\hat{i}+\hat{j}+\hat{k}

The given vectors are coplanar if,

A = u(B) + v(C)

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

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 \vec{a},\vec{b},\vec{c}  are non-coplanar vectors, prove that the given vectors are non-coplanar 2\vec{a}-\vec{b}+3\vec{c},\vec{a}+\vec{b}-2\vec{c},\vec{a}+\vec{b}-3\vec{c}

Solution:

The given vectors are D(2\vec{a}-\vec{b}+3\vec{c}),E(\vec{a}+\vec{b}-2\vec{c}),F(\vec{a}+\vec{b}-3\vec{c}) )

The given vectors are coplanar if,

D = u(E) + v(F)

2\vec{a}-\vec{b}+3\vec{c}=u(\vec{a}+\vec{b}-2\vec{c})+v(\vec{a}+\vec{b}-3\vec{c})
2\vec{a}-\vec{b}+3\vec{c}=(u+v)\vec{a}+(u+v)\vec{b}+(-2u-3v)\vec{c}

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 \vec{a},\vec{b},\vec{c}  are non-coplanar vectors, prove that the given vectors are non-coplanar \vec{a}+2\vec{b}+3\vec{c}, 2\vec{a}+\vec{b}+3\vec{c},\vec{a}+\vec{b}+\vec{c}

Solution:

The given vectors are D(\vec{a}+2\vec{b}+3\vec{c}),E(2\vec{a}+\vec{b}+3\vec{c}),F(\vec{a}+\vec{b}+\vec{c})

The given vectors are coplanar if,

D = u(E) + v(F)

\vec{a}+2\vec{b}+3\vec{c}=u(2\vec{a}+\vec{b}+3\vec{c})+v(\vec{a}+\vec{b}+\vec{c})
\vec{a}+2\vec{b}+3\vec{c}=(2u+v)\vec{a}+(u+v)\vec{b}+(3u+v)\vec{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 = -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 \vec{a},\vec{b},\vec{c}  given by \vec{a}=\hat{i}+2\hat{j}+3\hat{k},\vec{b}=2\hat{i}+\hat{j}+3\hat{k},\vec{c}=\hat{i}+\hat{j}+\hat{k}  are non-coplanar. Express vector \vec{d} =\vec{d} = 2\hat{i} -\hat{j}-3\hat{k}  as a linear combination of the vector \vec{a},\vec{b}, and\ \vec{c} .

Solution:

The given vectors are

\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\\ \vec{b}=2\hat{i}+\hat{j}+3\hat{k}\\ \vec{c}=\hat{i}+\hat{j}+\hat{k}

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

\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\\ \vec{b}=2\hat{i}+\hat{j}+3\hat{k}\\ \vec{c}=\hat{i}+\hat{j}+\hat{k}\\ \vec{d}=2\hat{i}-\hat{j}-3\hat{k}

The given vectors are coplanar if,

\vec{d}=\vec{a}x+\vec{b}y+\vec{c}z
2\hat{i}-\hat{j}-3\hat{k}=x(\hat{i}+2\hat{j}+3\hat{k})+y(2\hat{i}+\hat{j}+3\hat{k})+z(\hat{i}+\hat{j}+\hat{k})
2\hat{i}-\hat{j}-3\hat{k}=(x+2y+z)\hat{i}+(2x+y+z)\hat{j}+(3x+3y+z)\hat{k}

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, \vec{d} = \frac{-8}{3}\vec{a} + \frac{1}{3}\vec{b} + 4\vec{c}

Question 9. Prove that a necessary and sufficient condition for the three vectors \vec{a},\vec{b},and \ \vec{c}  to be coplanar is that these exist scalar l, m, n, not all zero simultaneously such that l\vec{a}+m\vec{b}+n\vec{c} = \vec{0}

Solution:

Given conditions: Let us considered \vec{a},\vec{b},and \ \vec{c} be three coplanar vectors.

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

Let,

c=x\vec{a} +y\vec{b}
x\vec{a}+y\vec{b}-\vec{c}=0

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

From above,

l\vec{a}+m\vec{b}+n\vec{c}=0
n\vec{c}=-l\vec{a}-m\vec{b}
\vec{c}=(-l/n)\vec{a}+(-m/n)\vec{b}

Hence, \vec{c}  is a linear combination of two vectors \vec{a} \ and \ \vec{b} .

Hence proved that \vec{a},\vec{b},and \ \vec{c}  are coplanar vectors.

Question 10. Show that the four points A, B, C, and D with position vectors \vec{a},\vec{b},\vec{c}, and \ \vec{d}  respectively are coplanar if and only if 3\vec{a}-2\vec{b}+\vec{c}-2\vec{d} = 0 .

Solution:

Given: A, B, C, D be four vectors with position vector \vec{a},\vec{b},\vec{c}, and \ \vec{d}  

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

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

x\vec{a} + y\vec{b} + z\vec{c} + u\vec{d} = 0

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

So, 3\vec{a}-2\vec{b}+\vec{c}-2\vec{d}=0

and x + y + z + u = 3 – 2 + 1 – 2 = 0

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

Let us considered 3\vec{a}-2\vec{b}+\vec{c}-2\vec{d}=0

3\vec{a}+\vec{c}=2\vec{b}+2\vec{d}

Now on dividing both side by sum of coefficient 4 

\frac{(3\vec{a}+\vec{c})}{4}=\frac{(2\vec{b}+2\vec{d})}{4}
\frac{(3\vec{a}+\vec{c})}{(3+1)}=\frac{(2\vec{b}+2\vec{d})}{(2+2)}

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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