3.4 Q-7

Exercise 3.4 — Question 7

Problem Statement

Find the value of $\lambda$ for which the vectors $\mathbf{a}=\hat{i}+2\hat{j}-3\hat{k}$, $\mathbf{b}=3\hat{i}-\hat{j}+2\hat{k}$, and $\mathbf{c}=\hat{i}+\lambda \hat{j}+\hat{k}$ are coplanar.

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Key Concept: Coplanarity Condition

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

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

In determinant form:

$\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=0$

This condition holds because coplanar vectors span a parallelepiped of zero volume.

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Solution

Step 1: Set up the determinant.

Write the components of each vector as rows of a $3\times 3$ matrix:

$\left|\begin{array}{ccc}1& 2& -3\\ 3& -1& 2\\ 1& \lambda & 1\end{array}\right|=0$

Step 2: Expand along the first row.

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

Step 3: Evaluate each $2\times 2$ determinant.

$=1\cdot [(-1)(1)-(2)(\lambda )]-2\cdot [(3)(1)-(2)(1)]+(-3)\cdot [(3)(\lambda )-(-1)(1)]$

$=1\cdot (-1-2\lambda )-2\cdot (3-2)+(-3)\cdot (3\lambda +1)$

$=(-1-2\lambda )-2(1)-3(3\lambda +1)$

$=-1-2\lambda -2-9\lambda -3$

$=-6-11\lambda$

Step 4: Set equal to zero and solve.

$-6-11\lambda =0$

$\lambda =-\frac{6}{11}$

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Answer

$\boxed{\lambda =-\frac{6}{11}}$

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Summary of Method