Vector Product (Part-6) | S N Dey | Class 12

by

In the previous article , we have solve the solutions of Ex-2B of Vector Product chapter of S N Dey mathematics class 12 book of Chhaya maths. In this article, we will discuss the solutions of Ex-2B of Product of two vectors chapter.

vector product, S N Dey mathematics class 12
Vector Product | S N Dey mathematics class 12 Solutions of Ex-2A

1(i)~~ If ~~\vec{a}=\hat{i}+\hat{j},~~ \vec{b}=\hat{i}-\hat{j}~~ and ~~ \vec{c}=5\hat{i}+2\hat{j} +3\hat{k},~~ find the value of ~~[\vec{b}~~\vec{c}~~\vec{a} ] .

Solution.

~~ [\vec{b}~~\vec{c}~~\vec{a}]\\= \begin{vmatrix} 1 & -1 & 0 \\ 5 & 2 & 3 \\ 1& 1& 0 \\ \end{vmatrix} \\ = 1\begin{vmatrix} 2& 3 \\ 1 & 0 \\ \end{vmatrix}+1\begin{vmatrix} 5 & 3 \\ 1 & 0 \\ \end{vmatrix} \\= (0-3)+(0-3)\=-6~~\text{(ans.)}

(ii)~~ If ~~ \vec{\alpha}=\hat{i}-2\hat{j} +3\hat{k}, ~~ \vec{\beta}=2\hat{i}-3\hat{j} +\hat{k}~~ and ~~ \vec{\gamma}=3\hat{i}+\hat{j} -2\hat{k},~~ find ~~ \vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma}).

Solution.

\vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma}) \\=[\vec{\alpha} ~~ \vec{\beta}~~\vec{\gamma}] \\= \begin{vmatrix} 1& -2 &3 \\ 2& -3&1 \\ 3&1 &-2 \\ \end{vmatrix} \\= 1(6-1)+2(-4-3)+3(2+9)\\=5-14+33\\=24~~\text{(ans.)}

(iii)~~ If ~~\vec{\alpha}=-\hat{i}+2\hat{j} +\hat{k},~~\vec{b}=3\hat{i}+\hat{j} +2\hat{k}~~ and ~~\vec{c}= 2\hat{i}+\hat{j} +3\hat{k},~~ find ~[\vec{c}~~\vec{a}~~\vec{b}].

Solution.

[\vec{c}~~\vec{a}~~\vec{b}] \\= \begin{vmatrix} 2& 1 & 3 \\ -1 & 2 & 1 \\ 3& 1 & 2 \\ \end{vmatrix} \\= 2\begin{vmatrix} 2& 1 \\ 1& 2 \\ \end{vmatrix}-1\begin{vmatrix} \\ -1 & 1 \\ 3& 2 \\ \end{vmatrix}+3\begin{vmatrix} -1& 2 \\ 3 & 1 \\ \end{vmatrix}\\=2(4-1)-1(-2-3)+3(-1-6) \\= 2 \times 3+5+3 \times (-7) \\=6+5-21\\=-10~~\text{(ans.)}

2.~~ The vectors which determine the sides of the parallelopiped are given below ; In each case find the volume of the parallelopiped .

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

Solution.

let ~~\vec{a}=\hat{i}+\hat{j} +\hat{k},~~\vec{b}=\hat{k},~\vec{c}=3\hat{i}-\hat{j} +2\hat{k}.

[\vec{c}~~\vec{a}~~\vec{b}] \\= \begin{vmatrix} 1& 1 &1 \\ 0&0 &1 \\ 3&-1 &2 \\ \end{vmatrix} \\ = -1 \begin{vmatrix} 1 & 1 \\ 3 &-1 \\ \end{vmatrix} \\= -1(-1-3) \\=4.

So, the volume of the parallelopiped is :

=| [\vec{c}~~\vec{a}~~\vec{b}]|=|4|=4~~\text{(cubic unit)}

(ii)~~ 2\hat{i}-\hat{j} +\hat{k},~~ \hat{i}+2\hat{j} -3\hat{k},~~ 3\hat{i}-4\hat{j} +5\hat{k}.

Solution.

Let ~~\vec{a}=2\hat{i}-\hat{j} +\hat{k},~~\vec{b}= \hat{i}+2\hat{j} -3\hat{k},~~\vec{c}= 3\hat{i}-4\hat{j} +5\hat{k}.

[\vec{a}~~\vec{b}~~\vec{c}] \\= \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & -4& 5 \\ \end{vmatrix}\\=2\begin{vmatrix} 2& -3 \\ -4 & 5 \\ \end{vmatrix}+1\begin{vmatrix} 1 & -3 \\ 3 & 5 \\ \end{vmatrix}+1 \begin{vmatrix} 1 & 2 \\ 3 & -4 \\ \end{vmatrix}\\=2(10-12)+1(5+9)+1(-4-6)\\=-4+14-10\\=0

So, the volume of the parallelopiped is :

= |[\vec{a}~~\vec{b}~~\vec{c}]|=|0|=0 ~~\text{cubic unit}.

(iii)~~\hat{i}+\hat{j} +\hat{k},~~ \hat{i}+2\hat{j} +2\hat{k},~~ \hat{i}-2\hat{j} +4\hat{k}.

Solution.

Let \vec{a}= \hat{i}+\hat{j} +\hat{k}, ~~\vec{b}=\hat{i}+2\hat{j} +2\hat{k},~~\vec{c}=\hat{i}-2\hat{j} +4\hat{k}.

[\vec{a}~~\vec{b}~~\vec{c}] \\= \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & -2& 4 \\ \end{vmatrix} \\= 1\begin{vmatrix} 2& 2 \\ -2 & 4 \\ \end{vmatrix}-1\begin{vmatrix} 1 & 2 \\ 1 & 4 \\ \end{vmatrix}+1 \begin{vmatrix} 1 & 2 \\ 1 & -2 \\ \end{vmatrix}\\=1(8+4)-1(4-2)+1(-2-2)\\=12-2-4\\=6

So, the volume of the parallelopiped is :

=|[\vec{a}~~\vec{b}~~\vec{c}]|=|6|=6 ~~\text{cubic unit.}

3.~~ In each of the following show that the given vectors are coplanar.

(i)~~~ 4\hat{i}+2\hat{j} +\hat{k},~~ 2\hat{i}-\hat{j} +3\hat{k},~~ 8\hat{i}+7\hat{k}.

Solution.

Let ~~\vec{p}=4\hat{i}+2\hat{j} +\hat{k},~~\vec{q}= 2\hat{i}-\hat{j} +3\hat{k},~~\vec{r}= 8\hat{i}+7\hat{k}. Then,

\vec{p} \cdot (\vec{q} \times \vec{r}) \\= \begin{vmatrix} 4& 2 & 1 \\ 2& -1 & 3 \\ 8& 0 & 7 \\ \end{vmatrix}\\=8 \begin{vmatrix} 2 & 1 \\ -1 &3 \\ \end{vmatrix}+7\begin{vmatrix} 4 & 2 \\ 2 & -1 \\ \end{vmatrix}\\=8(6+1)+7(-4-4)\\=56-56\\=0

Since ~~\vec{p} \cdot (\vec{q} \times \vec{r})=0,~~ hence the given vectors are coplanar.

(ii)~~ \vec{a} =\hat{i}+\hat{j} -6\hat{k},~~ \vec{b}=\hat{i}+3\hat{j} +4\hat{k}, ~~ \vec{c}=2\hat{i}+5\hat{j}+3\hat{k}.

Solution.

\vec{a} \cdot (\vec{b} \times \vec{c}) \\= \begin{vmatrix} 1& 1 & -6 \\ 1& 3 & 4 \\ 2& 5 & 3 \\ \end{vmatrix} \\= 1 \begin{vmatrix} 3 & 4 \\ 5 &3 \\ \end{vmatrix}- 1 \begin{vmatrix} 1 & 4 \\ 2 &3 \\ \end{vmatrix}-6\begin{vmatrix} \\ 1 & 3 \\ 2 & 5 \\ \end{vmatrix}\\=1(9-20)-1(3-8)-6(5-6)\\=-11+5+6\\=0

Since, ~~\vec{a} \cdot (\vec{b} \times \vec{c})=0,~~ hence the given vectors are coplanar.

(iii)~~ \vec{a}-2\vec{b}+\vec{c},~~ 2\vec{a}+\vec{b}-2\vec{c} ,~~ 3\vec{a}-\vec{b}-\vec{c}.

Solution.

(2\vec{a}+\vec{b}-2\vec{c} )\times (3\vec{a}-\vec{b}-\vec{c})\\= 6(\vec{a} \times \vec{a})-2(\vec{a} \times \vec{b})-2(\vec{a} \times \vec{c})+3(\vec{b} \times \vec{a})-(\vec{b} \times \vec{b})-(\vec{b}\times \vec{c})\\~~-6(\vec{c} \times \vec{a})+2(\vec{c} \times \vec{b})+2(\vec{c} \times \vec{c})\\= -2(\vec{a} \times \vec{b})+2(\vec{c} \times \vec{a})-3(\vec{a} \times \vec{b})-(\vec{b} \times \vec{c})-6(\vec{c} \times \vec{a})-2(\vec{b} \times \vec{c}) \\= -5(\vec{a} \times \vec{b})-3(\vec{b} \times \vec{c})-4(\vec{c} \times \vec{a})

(\vec{a}-2\vec{b}+\vec{c}) \cdot [(2\vec{a}+\vec{b}-2\vec{c} )\times (3\vec{a}-\vec{b}-\vec{c})]\\= (\vec{a}-2\vec{b}+\vec{c}) \cdot [-5(\vec{a} \times \vec{b})-3(\vec{b} \times \vec{c})-4(\vec{c} \times \vec{a})] \\= -3 [\vec{a}~~\vec{b}~~\vec{c}]+8[\vec{b}~~\vec{c}~~\vec{a}]-5 [\vec{c}~~\vec{a}~~\vec{b}]~~[*]\\=0 \rightarrow(1)

Hence, from the result obtained from ~(1),~~ we can conclude that the given vectors are coplanar.

Note[*] : ~~~ \vec{a} \cdot (\vec{a} \times \vec{b})=\vec{a} \cdot (\vec{c} \times \vec{a})=\vec{b} \cdot(\vec{a} \times \vec{b})=\cdots =0

4.~~ If ~~\vec{a}=-2\hat{i}-2\hat{j} +4\hat{k},~~ \vec{b}=-2\hat{i}+4\hat{j} -2\hat{k},~~ \vec{c}=4\hat{i}-2\hat{j} -2\hat{k},~~ find ~~\vec{a}\cdot (\vec{b} \times \vec{c})~~ and interpret the result.

Solution.

\vec{a}\cdot (\vec{b} \times \vec{c})\\=\begin{vmatrix} -2& -2 &4 \\ -2 &4 &-2\\ 4& -2 &-2 \\ \end{vmatrix} \\= -2\begin{vmatrix} 4 & -2\\ -2 & -2 \\ \end{vmatrix}+2\begin{vmatrix} -2&-2 \\ 4&-2 \\ \end{vmatrix}+4\begin{vmatrix} -2&4 \\ 4 &-2 \\ \end{vmatrix}\\=-2(-8-4)+2(4+8)+4(4-16)\\=24+24-48\\=0

Since, ~~\vec{a}\cdot (\vec{b} \times \vec{c})=0,~~ hence the given vectors are coplanar.

5.~~ If the vectors ~~x\hat{i}-4\hat{j} +5\hat{k},~~ \hat{i}+2\hat{j} +\hat{k}~~ and ~~ 2\hat{i}-\hat{j} +\hat{k}~~ are coplanar, find the value of ~x.

Solution.

let ~~\vec{a}=x\hat{i}-4\hat{j} +5\hat{k} ,~~ \vec{b}=\hat{i}+2\hat{j} +\hat{k},~~ \vec{c}=2\hat{i}-\hat{j} +\hat{k}.

Since ~~\vec{a},~~\vec{b},~~\vec{c}~~ are coplanar, ~~ \vec{a} \cdot (\vec{b} \times \vec{c})=0.

\therefore~ \vec{a} \cdot (\vec{b} \times \vec{c})=0=0 \\ \text{or,}~~ \begin{vmatrix} x& -4 & 5 \\ 1& 2 & 1 \\ 2& -1 & 1 \\ \end{vmatrix}=0 \\ \text{or,}~~ x(2+1)+4(1-2)+5(-1-4)=0 \\ \text{or,}~~ 3x-29=0 \\ \text{or,}~~ x=\frac{29}{3} ~~ \text{(ans.)}

6.~~ If the vectors ~~ \vec{a}=2\hat{i}-\hat{j} +\hat{k},~~ \vec{b}=\hat{i}+2\hat{j} -3\hat{k}~~ and ~~ \vec{c}=3\hat{i}+\lambda \hat{j} +5\hat{k}~~ are coplanar , find the value of ~~\lambda.

Solution.

If the vectors ~~\vec{a},~~\vec{b},~~\vec{c}~~ are coplanar, then

\vec{a} \cdot (\vec{b} \times \vec{c})=0 \\ \text{or,}~~\begin{vmatrix} 2& -1 &1 \\ 1& 2 & -3 \\ 3& \lambda & 5 \\ \end{vmatrix}=0 \\ \text{or,}~~2(10+3\lambda)+1(5+9)+1(\lambda-6)=0 \\ \text{or,}~~ 20+6\lambda+14+\lambda-6=0 \\ \text{or,}~~ 28+7\lambda=0 \\ \text{or,}~~ 7\lambda=-28 \\ \text{or,}~~ \lambda=-\frac{28}{7}=-4~~\text{(ans.)}

To download full solution PDF of Products of Two Vectors (S N De) , click here .

7.~~ The position vectors of four points ~A,B,C~ and ~D~ are given below. In each case, using vector method prove that the four points ~A,B,C~ and ~D~ are coplanar.

(i)~~ 6\hat{i}-4\hat{j} +10\hat{k}~;~~ -5\hat{i}+3\hat{j} -10\hat{k}; ~~4\hat{i}-6\hat{j} -10\hat{k};~~ 2\hat{j} +10\hat{k}.

Solution.

\vec{AB} \\= (-5\hat{i}+3\hat{j} -10\hat{k})-(6\hat{i}-4\hat{j} +10\hat{k}) \\= -11\hat{i}+7\hat{j} -20\hat{k},\\~~ \\~\vec{BC} \\= (4\hat{i}-6\hat{j} -10\hat{k})-(-5\hat{i}+3\hat{j} -10\hat{k}) \\= 9\hat{i}-9\hat{j},\\~~ \\~~ \vec{CD}\\=(2\hat{j} +10\hat{k})-(4\hat{i}-6\hat{j} -10\hat{k}) \\= -4\hat{i}+8\hat{j}+20\hat{k}.

[\vec{AB}~~\vec{BC}~~\vec{CD}]\\= \begin{vmatrix} -11& 7 &-20 \\ 9&-9 &0 \\ -4& 8 &20 \\ \end{vmatrix}\\=\begin{vmatrix} -15& 15&0 \\ 9 & -9 &0 \\ -4& 8 & 20 \\ \end{vmatrix} ~~[~R_1' \rightarrow R_1+R_3~]\\= 20(15 \times 9-15\times 9)\\=0

\therefore~ \vec{AB},~\vec{BC},~\vec{CD}~~ are coplanar and so, the four points ~~A,~B,~C~~ and ~~D~~ are coplanar.

(ii)~~ 4\hat{i}+8\hat{j} +12\hat{k}~;~ 2\hat{i}+4\hat{j} +6\hat{k};~ 3\hat{i}+5\hat{j} +4\hat{k};~ 5\hat{i}+8\hat{j} +5\hat{k}

Solution.

\vec{AB}\\= (2\hat{i}+4\hat{j} +6\hat{k})-(4\hat{i}+8\hat{j} +12\hat{k}) \\= -2\hat{i}-4\hat{j} -6\hat{k},\\~~\\~~ \vec{BC}\\=(3\hat{i}+5\hat{j} +4\hat{k})-(2\hat{i}+4\hat{j} +6\hat{k})\\=\hat{i}+\hat{j} -2\hat{k},\\~~\\~~ \vec{CD}\\=(5\hat{i}+8\hat{j} +5\hat{k})-(3\hat{i}+5\hat{j} +4\hat{k})\\=2\hat{i}+3\hat{j} +\hat{k}.

[ \vec{AB}~~\vec{BC}~~\vec{CD}]\\= \begin{vmatrix} -2& -4 &-6 \\ 1&1 &-2 \\ 2& 3 &1 \\ \end{vmatrix}\\=-2(1+6)+4(1+4)-6(3-2)\\=-2\times 7+20-6\\=-14+20-6\\=0

\therefore~~\vec{AB},~\vec{BC},~\vec{CD}~~ are coplanar and so, the four points ~A,B,C~ and ~D~ are coplanar.

8.~~ If the vectors ~~-4\hat{i}-6\hat{j} -2\hat{k},~~-\hat{i}+4\hat{j} +3\hat{k}~~ and ~~-8\hat{i}-\hat{j} +\lambda\hat{k}~~ are coplanar, then find the value of ~~\lambda.

Solution.

let ~~\vec{a}=-4\hat{i}-6\hat{j} -2\hat{k},~~\vec{b}=-\hat{i}+4\hat{j} +3\hat{k}~~ and ~~\vec{c}=-8\hat{i}-\hat{j} +\lambda\hat{k}

If the vectors ~~\vec{a},~\vec{b}~~ and ~~\vec{c}~~ are coplanar,

\vec{a} \cdot (\vec{b} \times \vec{c})=0 \\ \text{or,}~~ \begin{vmatrix} -4& -6 & -2 \\ -1& 4 &3 \\ -8& -1 & \lambda \\ \end{vmatrix} =0 \\ \text{or,}~~ -4(4\lambda+3)+6(-\lambda+24)-2(1+32)=0 \\ \text{or,}~~ -16\lambda-12-6\lambda+144-66=0 \\ \text{or,}~~ -22\lambda+66=0 \\ \text{or,}~~ -22\lambda=-66 \\ \text{or,}~~\lambda= \frac{-66}{-22}=3~~\text{(ans.)}

9.~~ If the vectors ~~\vec{a}=2\hat{i}-\lambda\hat{j} +3\hat{k},~~ \vec{b}=3\hat{i}+2\hat{j} -\mu\hat{k}~~ and ~~ \vec{c}=\hat{i}+\hat{j} +\hat{k}~~ are coplanar, find ~\mu~ in terms of ~\lambda.

Solution.

If the vectors ~~\vec{a},~\vec{b},~\vec{c}~~ are coplanar,

\vec{a} \cdot (\vec{b} \times \vec{c})=0 \\ \text{or,}~~ \begin{vmatrix} 2& -\lambda & 3 \\ 3& 2 &-\mu \\ 1& 1 & 1 \\\end{vmatrix} =0 \\ \text{or,}~~2(2+\mu)+\lambda(3+\mu)+3(3-2)=0 \\ \text{or,}~~ 4+2\mu+3\lambda+\lambda \mu+3=0 \\ \text{or,}~~ \mu(2+\lambda)+(7+3\lambda)=0 \\ \text{or,}~~ \mu(2+\lambda)=-(7+3\lambda) \\ \text{or,}~~ \mu=-\frac{7+3\lambda}{2+\lambda}~~\text{(ans.)}

10.~~ Prove that,

(i)~~ (\vec{a}+\vec{b}) \cdot {(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})}=2\vec{a}\cdot (\vec{b} \times \vec{c})

Solution.

(\vec{a}+\vec{b}) \cdot {(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})} \\= (\vec{a}+\vec{b}) \cdot (\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c}\times \vec{c}+\vec{c}\times \vec{a})\\= (\vec{a}+\vec{b}) \cdot (\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+0+\vec{c}\times \vec{a})\\= (\vec{a}+\vec{b}) \cdot (\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c}\times \vec{a}) \\= [\vec{a}~~\vec{b}~~\vec{c}] +[\vec{a}~~\vec{b}~~\vec{a}] +[\vec{a}~~\vec{c}~~\vec{a}] +[\vec{b}~~\vec{b}~~\vec{c}] +[\vec{b}~~\vec{b}~~\vec{a}] +[\vec{b}~~\vec{c}~~\vec{a}] \\=[\vec{a}~~\vec{b}~~\vec{c}] +0+0+0+0+[\vec{a}~~\vec{b}~~\vec{c}] \\=2[\vec{a}~~\vec{b}~~\vec{c}] \\=2\vec{a} \cdot (\vec{b} \times \vec{c})~~\text{(proved)}

(ii)~~ \vec{a} \cdot {\vec{b} \times (\vec{c}+\vec{d}) }=\vec{a} \cdot (\vec{b} \times \vec{c})+\vec{a} \cdot (\vec{b} \times \vec{d})

Solution.

\vec{a} \cdot {\vec{b} \times (\vec{c}+\vec{d})}\\=\vec{a} \cdot {\vec{b} \times \vec{c}+\vec{b} \times \vec{d}}\\=\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot (\vec{b} \times \vec{d}) ~~\text{(proved)}

11.~~ \vec{\alpha}=\hat{i}+\hat{j} +\hat{k}, ~~ \vec{\beta}=\hat{i}+\hat{j} +\hat{k},~~ \vec{\gamma}=\hat{i}+\hat{j} +\hat{k}~~ and ~~[\vec{\alpha}~~\vec{\beta}~~\vec{\gamma}] =-10,~~ then find the value of ~~\lambda.

Solution.

[\vec{\alpha}~~\vec{\beta}~~\vec{\gamma}] =-10 \\ \text{or,}~~ \begin{vmatrix} \lambda & 1 & 3 \\ -1 &2 &1 \\ 3& 1&2 \\ \end{vmatrix} =-10 \\ \text{or,}~~ \lambda(4-1)-1(-2-3)+3(-1-6)=-10 \\ \text{or,}~~ 3\lambda+5-21=-10 \\ \text{or,}~~ 3\lambda-16=-10 \\ \text{or,}~~3\lambda=-10+16 \\ \text{or,}~~ \lambda=\frac 63=2~~\text{(ans.)}

12.~~ If the vectors ~~ a\hat{i}+a\hat{j} +c\hat{k},~~ \hat{i} +\hat{k}~~ and ~~c\hat{i}+c\hat{j} +b\hat{k}~~ be coplanar, show that ~~c^2=ab.

Solution.

let ~~\vec{\alpha}= a\hat{i}+a\hat{j} +c\hat{k},~~ \vec{\beta}=\hat{i} +\hat{k},~~ \vec{\gamma}=c\hat{i}+c\hat{j} +b\hat{k}.

If ~~\vec{\alpha},~~\vec{\beta},~~\vec{\gamma}~~ are coplanar, then

\vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma})=0 \\ \text{or,}~~\begin{vmatrix} a& a & c \\ 1& 0 & 1 \\ c& c &b \\ \end{vmatrix}=0 \\ \text{or,}~~a(0-c)-a(b-c)+c(c-0)=0 \\ \text{or,}~~ -ac-ab+ac+c^2=0 \\ \text{or,}~~ -ab+c^2=0 \\ \text{or,}~~ c^2=ab~ ~ \text{(showed)}

13.~~ Let ~~\vec{a}= \hat{i}+\hat{j} +\hat{k},~~ \vec{b}=\hat{i}~~ and ~~\vec{c}=c_1\hat{i}+c_2\hat{j} +c_3\hat{k}~~ then, if ~~c_1=1~~ and ~~c_2=2,~~ find ~~c_3~~ which makes ~~\vec{a},~~\vec{b}~~ and ~~\vec{c}~~ coplanar.

Solution.

Since ~~\vec{a},~~\vec{b},~~\vec{c}~~ are coplanar, then

\vec{a} \cdot (\vec{b} \times \vec{c})=0 \\ \text{or,}~~ \begin{vmatrix} 1& 1 & 1 \\ 1& 0 & 0 \\ c_1& c_2 &c_3 \\ \end{vmatrix}=0 \\ \text{or,}~~ \begin{vmatrix} 1& 1 & 1 \\ 1& 0 & 0 \\ 1& 2 &c_3 \\\end{vmatrix}=0 \\ \text{or,}~~ -1 \begin{vmatrix} 1& 1 \\ 2& c_3 \\ \end{vmatrix}=0 \\ \text{or,}~~ -(c_3-2)=0 \\ \text{or,}~~ c_3=2~~\text{(ans.)}

14.~~ Find ~x~ such that the four points ~A(3,2,1),~ B(4,x,5),~ C(4,2,-2)~ and ~~D(6,5,-1)~~ are coplanar.

Solution.

let ~~\vec{AB}=\hat{i}+(x-2)\hat{j} +4\hat{k},~~ \vec{BC}=(2-x)\hat{j} -7\hat{k},~~ \vec{CD}=2\hat{i}+3\hat{j} +\hat{k}.

If the four points ~~A,~B,~C,~D~~ are coplanar, ~~\vec{AB},~~\vec{BC},~~\vec{CD}~~ are coplanar.

\therefore~~ \vec{AB} \cdot (\vec{BC} \times \vec{CD})=0 \\ \text{or,}~~ \begin{vmatrix} 1& x-2 & 4 \\ 0& 2-x &-7 \\ 2& 3 & 1 \\ \end{vmatrix}=0 \\ \text{or,}~~ 1(2-x+21)-(x-2)(0+14)+4(0-4+2x)=0 \\ \text{or,}~~ 23-x-14x+28-16+8x=0 \\ \text{or,}~~ 7x=35 \\ \text{or,}~~ x=\frac{35}{7}=5~~\text{(ans.)}

Post Views: 263

Leave a Comment