Question

Check if the points with position vectors $6\vec{a}-4\vec{b}+10\vec{c},-5\vec{a}+3\vec{b}+10\vec{c},4\vec{a}-6\vec{b}-10\vec{c}\text{ and }2\vec{b}+10\vec{c}$ are coplanar if $\vec{a},\vec{b},\vec{c}$ are non-coplanar vectors.

Answer 1

In this question, we are given position vectors of four points and we have to check if they are coplanar or not. For this, we will first consider these points as P, Q, R, S and their position vectors as $\overrightarrow{OP},\overrightarrow{OQ},\overrightarrow{OR},\overrightarrow{OS}$. Then, we will find $\overrightarrow{PQ},\overrightarrow{PR},\overrightarrow{PS}$. After this, we will put the coefficient of $\vec{a},\vec{b},\vec{c}$ in the determinant form and find its value to check coplanarity. For three vectors ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},{{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\text{ and }{{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ the determinant $\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\ \end{matrix} \right|$ should be equal to zero for the vectors to be coplanar.

Complete step-by-step solution:
Here, we are given points as P, Q, R, S
Their position vectors are

& \overrightarrow{OP}=6\vec{a}-4\vec{b}+10\vec{c} \\\ & \overrightarrow{OQ}=-5\vec{a}+3\vec{b}+10\vec{c} \\\ & \overrightarrow{OR}=4\vec{a}-6\vec{b}-10\vec{c} \\\ & \overrightarrow{OS}=2\vec{b}+10\vec{c} \\\ \end{aligned}$$ As we are given four vectors, we have to change them to three vectors by finding vectors $\overrightarrow{PQ},\overrightarrow{PR},\overrightarrow{PS}$. We know, $$\overrightarrow{PQ}=\text{Position vector of }\overrightarrow{OQ}-\text{Position vector of }\overrightarrow{OP}$$ Therefore, $$\begin{aligned} & \overrightarrow{PQ}=\left( -5\vec{a}+3\vec{b}+10\vec{c} \right)-\left( 6\vec{a}-4\vec{b}+10\vec{c} \right) \\\ & \Rightarrow -5\vec{a}+3\vec{b}+10\vec{c}-6\vec{a}+4\vec{b}-10\vec{c} \\\ & \Rightarrow -11\vec{a}+7\vec{b} \\\ \end{aligned}$$ Similarly, $$\begin{aligned} & \overrightarrow{PR}=\text{Position vector of }\overrightarrow{OR}-\text{Position vector of }\overrightarrow{OP} \\\ & \overrightarrow{PR}=\left( 4\vec{a}-6\vec{b}-10\vec{c} \right)-\left( 6\vec{a}-4\vec{b}+10\vec{c} \right) \\\ & \Rightarrow 4\vec{a}-6\vec{b}-10\vec{c}-6\vec{a}+4\vec{b}-10\vec{c} \\\ & \Rightarrow -2\vec{a}-2\vec{b}-20\vec{c} \\\ \end{aligned}$$ And $$\begin{aligned} & \overrightarrow{PS}=\text{Position vector of }\overrightarrow{OS}-\text{Position vector of }\overrightarrow{OP} \\\ & \overrightarrow{PS}=\left( 2\vec{b}+10\vec{c} \right)-\left( 6\vec{a}-4\vec{b}+10\vec{c} \right) \\\ & \Rightarrow 2\vec{b}+10\vec{c}-6\vec{a}+4\vec{b}-10\vec{c} \\\ & \Rightarrow -6\vec{a}+6\vec{b} \\\ \end{aligned}$$ Now let us check if the vectors obtained are coplanar or not by taking the coefficient of vectors obtained as rows of the determinant matrix, hence, the determinant matrix becomes $$\left| \begin{matrix} -11 & 7 & 0 \\\ -2 & -2 & -20 \\\ -6 & 6 & 0 \\\ \end{matrix} \right|$$ Let us calculate determinant of this matrix using third column, we get: $$\begin{aligned} & \text{Determinant}=\left( 0 \right)\left( -12+12 \right)-\left( -20 \right)\left( -66+42 \right)+0 \\\ & \Rightarrow 20\left( 24 \right)=480 \\\ \end{aligned}$$ **Since, determinant of coefficient of vectors obtained is not equal to zero, hence, given vectors are not coplanar.** **Note:** While converting four vectors to three vectors, we can take vectors with respect to $\vec{R},\vec{S}\text{ or }\vec{Q}$ also, which means, we can take $\overrightarrow{RP},\overrightarrow{RQ},\overrightarrow{RS}\Rightarrow \overrightarrow{SP},\overrightarrow{SQ},\overrightarrow{SR}\Rightarrow \overrightarrow{QP},\overrightarrow{QR},\overrightarrow{QS}$ vectors also. Students should note that, while taking determinant, take care of the sign for every element. Sign is determined as ${{\left( -1 \right)}^{\text{No}\text{.of rows+No}\text{.of columns}}}$. For example, for element in second row and third column, the sign will be ${{\left( -1 \right)}^{2+3}}={{\left( -1 \right)}^{5}}=-1$. Also, determinant can be taken using any row or any column. While finding $\overrightarrow{AB}$ always subtract position vector of $\overrightarrow{A}$ from position vector of $\overrightarrow{A}$ only and don't change the order. $\overrightarrow{AB}\text{ and }\overrightarrow{BA}$ represent two different vectors.