Key Concepts and Formulas

1. Volume of a Parallelopiped: The volume $V$ of a parallelopiped with coterminous edges given by vectors $\overrightarrow u$, $\overrightarrow v$, and $\overrightarrow w$ is the absolute value of their scalar triple product (STP): $V = |(\overrightarrow u \times \overrightarrow v) \cdot \overrightarrow w| = \left| \det \begin{pmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{pmatrix} \right|$
2. Angle Between Two Vectors: For two non-zero vectors $\overrightarrow a$ and $\overrightarrow b$, the cosine of the angle $\theta$ between them is given by: $\cos \theta = \frac{\overrightarrow a \cdot \overrightarrow b}{|\overrightarrow a| |\overrightarrow b|}$ where $\overrightarrow a \cdot \overrightarrow b = a_x b_x + a_y b_y + a_z b_z$ and $|\overrightarrow a| = \sqrt{a_x^2 + a_y^2 + a_z^2}$.

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Step-by-Step Solution

1. Identify Given Information and Set Up the Volume Equation

We are given the coterminous edges of a parallelopiped:

• $\overrightarrow u = \widehat i + \widehat j + \lambda \widehat k$
• $\overrightarrow v = \widehat i + \widehat j + 3\widehat k$
• $\overrightarrow w = 2\widehat i + \widehat j + \widehat k$

The volume of the parallelopiped is given as $V = 1$ cubic unit.

• Why this step? The problem provides the volume and the vectors with an unknown parameter $\lambda$. We must use the volume information to find the possible values of $\lambda$.

Using the formula for the volume of a parallelopiped: $V = \left| \det \begin{pmatrix} 1 & 1 & \lambda \\ 1 & 1 & 3 \\ 2 & 1 & 1 \end{pmatrix} \right|$ Since $V=1$: $\left| \det \begin{pmatrix} 1 & 1 & \lambda \\ 1 & 1 & 3 \\ 2 & 1 & 1 \end{pmatrix} \right| = 1$

2. Calculate the Determinant and Solve for $\lambda$

Let's expand the determinant: $\det \begin{pmatrix} 1 & 1 & \lambda \\ 1 & 1 & 3 \\ 2 & 1 & 1 \end{pmatrix} = 1(1 \cdot 1 - 3 \cdot 1) - 1(1 \cdot 1 - 3 \cdot 2) + \lambda(1 \cdot 1 - 1 \cdot 2)$ $= 1(1 - 3) - 1(1 - 6) + \lambda(1 - 2)$ $= 1(-2) - 1(-5) + \lambda(-1)$ $= -2 + 5 - \lambda$ $= 3 - \lambda$

Now, we use the volume condition: $|3 - \lambda| = 1$ This equation leads to two possibilities:

• Case 1: $3 - \lambda = 1 \implies \lambda = 3 - 1 \implies \lambda = 2$
• Case 2: $3 - \lambda = -1 \implies \lambda = 3 - (-1) \implies \lambda = 4$

So, the possible values for $\lambda$ are $2$ and $4$.

3. Calculate $\cos \theta$ for Each Value of $\lambda$

We need to find the cosine of the angle $\theta$ between $\overrightarrow u$ and $\overrightarrow w$. The formula is $\cos \theta = \frac{\overrightarrow u \cdot \overrightarrow w}{|\overrightarrow u| |\overrightarrow w|}$.

First, let's find the magnitude of $\overrightarrow w$: $|\overrightarrow w| = |2\widehat i + \widehat j + \widehat k| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$

Now, we examine each case for $\lambda$:

Case 1: $\lambda = 2$

• Why this step? We have found two potential values for $\lambda$, and the problem asks for a specific value of $\cos \theta$. We must check the outcome for each $\lambda$.

If $\lambda = 2$, then $\overrightarrow u = \widehat i + \widehat j + 2\widehat k$. Calculate the dot product $\overrightarrow u \cdot \overrightarrow w$: $\overrightarrow u \cdot \overrightarrow w = (\widehat i + \widehat j + 2\widehat k) \cdot (2\widehat i + \widehat j + \widehat k) = (1)(2) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5$ Calculate the magnitude of $\overrightarrow u$: $|\overrightarrow u| = |\widehat i + \widehat j + 2\widehat k| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$ Now, calculate $\cos \theta$: $\cos \theta = \frac{\overrightarrow u \cdot \overrightarrow w}{|\overrightarrow u| |\overrightarrow w|} = \frac{5}{\sqrt{6} \cdot \sqrt{6}} = \frac{5}{6}$

Case 2: $\lambda = 4$

• Why this step? To ensure we cover all possibilities derived from the volume condition.

If $\lambda = 4$, then $\overrightarrow u = \widehat i + \widehat j + 4\widehat k$. Calculate the dot product $\overrightarrow u \cdot \overrightarrow w$: $\overrightarrow u \cdot \overrightarrow w = (\widehat i + \widehat j + 4\widehat k) \cdot (2\widehat i + \widehat j + \widehat k) = (1)(2) + (1)(1) + (4)(1) = 2 + 1 + 4 = 7$ Calculate the magnitude of $\overrightarrow u$: $|\overrightarrow u| = |\widehat i + \widehat j + 4\widehat k| = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$ Simplify $\sqrt{18}$: $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$. Now, calculate $\cos \theta$: $\cos \theta = \frac{\overrightarrow u \cdot \overrightarrow w}{|\overrightarrow u| |\overrightarrow w|} = \frac{7}{\sqrt{18} \cdot \sqrt{6}} = \frac{7}{\sqrt{108}}$ Simplify $\sqrt{108}$: $\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$. So, $\cos \theta = \frac{7}{6\sqrt{3}}$

4. Match the Result with the Given Options

We have found two possible values for $\cos \theta$: $\frac{5}{6}$ and $\frac{7}{6\sqrt{3}}$. Let's compare these with the given options: (A) ${7 \over {6\sqrt 3 }}$ (B) ${7 \over {6\sqrt 6 }}$ (C) ${5 \over 7}$ (D) ${5 \over {3\sqrt 3 }}$

Our calculated value $\frac{7}{6\sqrt{3}}$ matches option (A).

Common Mistakes & Tips

• Absolute Value: Always remember to take the absolute value of the scalar triple product when calculating the volume of a parallelopiped. Forgetting this can lead to missing one of the possible values for $\lambda$.
• Determinant Calculation: Double-check your determinant expansion to avoid arithmetic errors.
• Square Root Simplification: Be proficient in simplifying radicals (e.g., $\sqrt{18} = 3\sqrt{2}$, $\sqrt{108} = 6\sqrt{3}$) to match the format of the answer options.

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Summary

The problem required us to first use the given volume of the parallelopiped to find the possible values of the unknown parameter $\lambda$ by calculating the scalar triple product. We then used these values of $\lambda$ to determine the vector $\overrightarrow u$ and subsequently calculated the cosine of the angle between $\overrightarrow u$ and $\overrightarrow w$ using the dot product formula. One of the calculated values for $\cos \theta$ matched one of the provided options.

The final answer is $\boxed{{7 \over {6\sqrt 3 }}}$, which corresponds to option (A).