The volume of the parallelepiped whose coterminus edges are $7\hat i+\lambda\hat j-3\hat k,\ \hat i+2\hat j-\hat k,\ -3\hat i+7\hat j+5\hat k$ is $90$ cubic units. Find the value of $\lambda$.

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Accepted Answer

> [!TLDR]
>
> Expand the determinant symbolically in $\lambda$, then set its absolute value to the given volume $90$ — this gives two possible values since a volume constraint only fixes $|\det|$, not the sign of $\det$.

**Step 1. Set up and expand the determinant.**
$$\begin{vmatrix}7&\lambda&-3\1&2&-1\-3&7&5\end{vmatrix}=7\begin{vmatrix}2&-1\7&5\end{vmatrix}-\lambda\begin{vmatrix}1&-1\-3&5\end{vmatrix}+(-3)\begin{vmatrix}1&2\-3&7\end{vmatrix}.$$

**Step 2. Evaluate the minors.**
$\begin{vmatrix}2&-1\7&5\end{vmatrix}=10+7=17,\quad \begin{vmatrix}1&-1\-3&5\end{vmatrix}=5-3=2,\quad \begin{vmatrix}1&2\-3&7\end{vmatrix}=7+6=13.$

**Step 3. Combine.**
$$7(17)-\lambda(2)-3(13)=119-2\lambda-39=80-2\lambda.$$

**Step 4. Set $|80-2\lambda|=90$.** Either $80-2\lambda=90\Rightarrow\lambda=-5$, or $80-2\lambda=-90\Rightarrow\lambda=85$.
> [!ANSWER]
> $\lambda=\boxed{-5}$ or $\lambda=\boxed{85}$.

Q.The volume of the parallelepiped whose coterminus edges are $7\hat i+\lambda\hat j-3\hat k,\ \hat i+2\hat j-\hat k,\ -3\hat i+7\hat j+5\hat k$ is $90$ cubic units. Find the value of $\lambda$ .

Concept understanding — Scalar Triple Product and Coplanarity

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Q.The volume of the parallelepiped whose coterminus edges are 7i^+λj^−3k^, i^+2j^−k^, −3i^+7j^+5k^7\hat i+\lambda\hat j-3\hat k,\ \hat i+2\hat j-\hat k,\ -3\hat i+7\hat j+5\hat k7i^+λj^​−3k^, i^+2j^​−k^, −3i^+7j^​+5k^ is 909090 cubic units. Find the value of λ\lambdaλ.

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Q.The volume of the parallelepiped whose coterminus edges are $7\hat i+\lambda\hat j-3\hat k,\ \hat i+2\hat j-\hat k,\ -3\hat i+7\hat j+5\hat k$ is $90$ cubic units. Find the value of $\lambda$ .

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