Question 7 Exercise 3.5

Solutions of Question 7 of Exercise 3.5 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

For what value of $c$ the following vectors are coplanar $\vec{u}=\hat{i}+2 \hat{j}+3 \hat{k}$. $\vec{v}=2 \hat{i}-3 \hat{j}+4 \hat{k} \cdot \vec{w}=3 \hat{i}+\hat{j}+c \hat{k}$

The given vectors are coplanar, therefore $$\begin{align}\vec{u} \cdot \vec{v} \times \vec{w}&=0\\ \vec{u} \cdot \vec{v} \times \vec{w}&=0\\ \Rightarrow\left|\begin{array}{ccc}1 & 2 & 3 \\ 2 & -3 & 4 \\ 3 & 1 & c\end{array}\right|&=0\\ 1(-3 c-4)-2(2 c-12)+3(2+9)&=0\\ \Rightarrow-3 c-4-4 c+24+33&=0\\ \Rightarrow \quad-7 c+53&=0\\ \Rightarrow c&=\dfrac{53}{7}.\end{align}$$ which is required value of $c$ for which the given vectors become coplanar.

For what value of $c$ the following vectors are coplanar $\vec{u}=\hat{i}+\hat{j}-\hat{k}$. $\vec{v}=\hat{i}-2 \hat{j}+\hat{k}, \vec{w}=c \hat{i}+\hat{j}-c \hat{k}$.

The given vectors are coplanar, therefore $$\begin{align}\vec{u} \cdot \vec{v} \times \vec{w}&=0\\ \Rightarrow\left|\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -2 & 1 \\ c & 1 & -c \end{array}\right|&=0\\ (2 c-1)-(-c-c)-(1+2 c)&=0 \\ \Rightarrow 2 c-1+2 c-1-2 c&=0 \\ \Rightarrow 2 c-2&=0\\ \Rightarrow c&=1\end{align}$$ which is the required value of $c$ for which the given vectors become coplanar.

For what value of $c$ the following vectors are coplanar $\vec{u}=\hat{i}+\hat{j}+2 \hat{k}, \vec{v}=2 \hat{i}+3 \hat{j}+\hat{k}$. $\vec{n}=c \hat{i}+2 \hat{j}+6 \hat{k}$

Since the given vectors are coplanar, therefore $$\begin{align}\vec{u} \cdot \bar{v} \times \vec{w}&=0 \\ \Rightarrow\left|\begin{array}{lll} 1 & 1 & 2 \\ 2 & 3 & 1 \\ c & 2 & 6 \end{array}\right|&=0\\ 1(18-2)-1(12-c)+2(4-3 c)&=0 \\ 16-12+c+8-6 c&=0 \\ \Rightarrow -5 c+12&=0 \\ \Rightarrow c&=\dfrac{-12}{-5}=\dfrac{12}{5}\end{align}$$ which is the required value of $c$ for which the given vectors become coplanar.

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