第12类RD Sharma-第23章向量的代数–练习23.8

问题1.证明给定位置矢量的点是共线的：

(一世) $2\hat{i}+\hat{j}-\hat{k}, 3\hat{i}-2\hat{j}+\hat{k}, and \ \hat{i}+4\hat{j}-3\hat{k}$

解决方案：

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

解决方案：

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.

问题2(i)。使用矢量方法，证明A(6，-7，-1)，B(2，-3，1)和C(4，-5，0)是共线的。

解决方案：

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.

问题2(ii)。使用矢量方法，证明A(2，-1，3)，B(4，3，1)和C(3，1，2)是共线的。

解决方案：

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.

问题2(iii)。使用矢量方法，证明X(1、2、7)，Y(2、6、3)和Z(3、10，-1)是共线的。

解决方案：

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.

问题2(iv)。使用矢量方法，证明X(-3，-2，-5)，Y(1、2、3)和Z(3、4、7)是共线的。

解决方案：

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.

问题2(v)。使用矢量方法，证明X(2，-1，3)，Y(3，-5，1)和Z(-1、11、9)是共线的。

解决方案：

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

$\vec{Y}=3\hat{i}-5\hat{j}+\hat{k}$

$\vec{Z}=-\hat{i}+11\hat{j}+9\hat{k}$

$\overrightarrow{XY}$= 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.

问题3(i)。如果$\vec{a},\vec{b},\vec{c}$是非零，非共平面向量，证明向量$5\vec{a}+6\vec{b}+7\vec{c}, 7\vec{a}-8\vec{b}+9\vec{c}, and \ 3\vec{a}+20\vec{b}+5\vec{c}$是共面的。

解决方案：

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.

问题3(ii)。如果$\vec{a},\vec{b},\vec{c}$是非零，非共平面向量，证明向量$\vec{a}-2\vec{b}+3\vec{c}, -3\vec{b}+5\vec{c}, and \ -2\vec{a}+3\vec{b}-4\vec{c}$是共面的。

解决方案：

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.

问题4：证明四个点具有位置向量$6\hat{i}-7\hat{j}, 16\hat{i}-19\hat{j}-4\hat{k},3\hat{i}-6\hat{k},2\hat{i}-5\hat{j}+10\hat{k}$是共面的。

解决方案：

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.

问题5(i)。证明以下向量是共面的显示点$2\hat{i}-\hat{j}+\hat{k},\hat{i}-3\hat{j}-5\hat{k},3\hat{i}-4\hat{j}-4\hat{k}$

解决方案：

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.

问题5(ii)。证明以下向量是共面的显示点$\hat{i}+\hat{j}+\hat{k},2\hat{i}+3\hat{j}-\hat{k},-\hat{i}-2\hat{j}+2\hat{k}$

解决方案：

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.

问题6(i)。证明向量$3\hat{i}+\hat{j}-\hat{k},2\hat{i}-\hat{j}+7\hat{k},7\hat{i}-\hat{j}+23\hat{k}$是非共面的。

解决方案：

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.

问题6(ii)。证明向量$\hat{i}+2\hat{j}+3\hat{k},2\hat{i}+\hat{j}+3\hat{k},\hat{i}+\hat{j}+\hat{k}$是非共面的。

解决方案：

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.

问题7(i)。如果$\vec{a},\vec{b},\vec{c}$是非共面的向量，证明给定的向量是非共面的$2\vec{a}-\vec{b}+3\vec{c},\vec{a}+\vec{b}-2\vec{c},\vec{a}+\vec{b}-3\vec{c}$

解决方案：

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.

问题7(ii)。如果$\vec{a},\vec{b},\vec{c}$是非共面的向量，证明给定的向量是非共面的$\vec{a}+2\vec{b}+3\vec{c}, 2\vec{a}+\vec{b}+3\vec{c},\vec{a}+\vec{b}+\vec{c}$

解决方案：

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.

问题8：证明向量$\vec{a},\vec{b},\vec{c}$由$\vec{a}=\hat{i}+2\hat{j}+3\hat{k},\vec{b}=2\hat{i}+\hat{j}+3\hat{k},\vec{c}=\hat{i}+\hat{j}+\hat{k}$是非共面的。表达向量\ vec {d} = $\vec{d} = 2\hat{i} -\hat{j}-3\hat{k}$作为向量的线性组合$\vec{a},\vec{b}, and\ \vec{c}$ 。

解决方案：

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

问题9.证明三个向量的充要条件$\vec{a},\vec{b},and \ \vec{c}$共面的是，它们存在标量l，m，n，并非同时为零，因此$l\vec{a}+m\vec{b}+n\vec{c} = \vec{0}$

解决方案：

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.

问题10：证明带有位置矢量的四个点A，B，C和D $\vec{a},\vec{b},\vec{c}, and \ \vec{d}$当且仅当$3\vec{a}-2\vec{b}+\vec{c}-2\vec{d} = 0$ 。

解决方案：

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.