3.3 Q-17

Exercise 3.3 — Question 17

Problem Statement

Prove that the vectors $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=-2\hat{i}+3\hat{j}-4\hat{k}$, and $\vec{c}=\hat{i}-3\hat{j}+5\hat{k}$ are coplanar.

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Key Concept: Coplanar Vectors

Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if and only if their scalar triple product is zero:

$\vec{a}\cdot (\vec{b}\times \vec{c})=0$

This is because the scalar triple product gives the volume of the parallelepiped formed by the three vectors. If the volume is zero, the vectors lie in the same plane.

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Scalar Triple Product in Determinant Form

If $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$, then:

$\vec{a}\cdot (\vec{b}\times \vec{c})=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$

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Solution

Given: $\vec{a}=\hat{i}-2\hat{j}+3\hat{k},\vec{b}=-2\hat{i}+3\hat{j}-4\hat{k},\vec{c}=\hat{i}-3\hat{j}+5\hat{k}$

Compute the scalar triple product:

$\vec{a}\cdot (\vec{b}\times \vec{c})=\left|\begin{array}{ccc}1& -2& 3\\ -2& 3& -4\\ 1& -3& 5\end{array}\right|$

Expanding along the first row:

$=1\left|\begin{array}{cc}3& -4\\ -3& 5\end{array}\right|-(-2)\left|\begin{array}{cc}-2& -4\\ 1& 5\end{array}\right|+3\left|\begin{array}{cc}-2& 3\\ 1& -3\end{array}\right|$

First minor: $\left|\begin{array}{cc}3& -4\\ -3& 5\end{array}\right|=(3)(5)-(-4)(-3)=15-12=3$

Second minor: $\left|\begin{array}{cc}-2& -4\\ 1& 5\end{array}\right|=(-2)(5)-(-4)(1)=-10+4=-6$

Third minor: $\left|\begin{array}{cc}-2& 3\\ 1& -3\end{array}\right|=(-2)(-3)-(3)(1)=6-3=3$

Substituting back:

$=1(3)+2(-6)+3(3)$ $=3-12+9$ $=0$

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Conclusion

Since $\vec{a}\cdot (\vec{b}\times \vec{c})=0$, the three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar. $■$

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Summary of Method