Key Concepts and Formulas

• Length of Projection Vector: The length of the projection of vector $\overrightarrow P$ onto vector $\overrightarrow Q$ is given by $\frac{|\overrightarrow P \cdot \overrightarrow Q|}{|\overrightarrow Q|}$.
• Scalar Triple Product (STP): The scalar triple product of vectors $\overrightarrow A, \overrightarrow B, \overrightarrow C$ is $[\overrightarrow A, \overrightarrow B, \overrightarrow C] = \overrightarrow A \cdot (\overrightarrow B \times \overrightarrow C)$. It has the property $\overrightarrow A \cdot (\overrightarrow B \times \overrightarrow C) = (\overrightarrow A \times \overrightarrow B) \cdot \overrightarrow C$.
• Cross Product Properties: The cross product is anti-commutative, i.e., $\overrightarrow A \times \overrightarrow B = -(\overrightarrow B \times \overrightarrow A)$.
• Dot Product Property: The dot product of a vector with itself is the square of its magnitude, i.e., $\overrightarrow V \cdot \overrightarrow V = |\overrightarrow V|^2$.

Step-by-Step Solution

Step 1: Define the projection length 'l'

The problem states that $l$ is the length of the projection vector of $\overrightarrow b$ on $\overrightarrow a \times \overrightarrow c$. Using the formula for the length of a projection vector, we have: $l = \frac{|\overrightarrow b \cdot (\overrightarrow a \times \overrightarrow c)|}{|\overrightarrow a \times \overrightarrow c|}$ Our goal is to calculate the numerator and the denominator of this expression.

Step 2: Calculate the denominator $|\overrightarrow a \times \overrightarrow c|$

We are given $\overrightarrow a = \widehat i + \widehat j + \widehat k$ and $\overrightarrow c = \widehat j - \widehat k$. First, we compute the cross product $\overrightarrow a \times \overrightarrow c$: $\overrightarrow a \times \overrightarrow c = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix}$ Expanding the determinant: $\overrightarrow a \times \overrightarrow c = \widehat i((1)(-1) - (1)(1)) - \widehat j((1)(-1) - (1)(0)) + \widehat k((1)(1) - (1)(0))$ $\overrightarrow a \times \overrightarrow c = \widehat i(-1 - 1) - \widehat j(-1 - 0) + \widehat k(1 - 0)$ $\overrightarrow a \times \overrightarrow c = -2\widehat i + \widehat j + \widehat k$ Now, we find the magnitude of this vector: $|\overrightarrow a \times \overrightarrow c| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ This is the denominator of $l$.

Step 3: Evaluate the numerator $|\overrightarrow b \cdot (\overrightarrow a \times \overrightarrow c)|$

The numerator involves the scalar triple product $\overrightarrow b \cdot (\overrightarrow a \times \overrightarrow c)$. We can use the property $\overrightarrow X \cdot (\overrightarrow Y \times \overrightarrow Z) = (\overrightarrow X \times \overrightarrow Y) \cdot \overrightarrow Z$. Let $\overrightarrow X = \overrightarrow b$, $\overrightarrow Y = \overrightarrow a$, and $\overrightarrow Z = \overrightarrow c$. Then: $\overrightarrow b \cdot (\overrightarrow a \times \overrightarrow c) = (\overrightarrow b \times \overrightarrow a) \cdot \overrightarrow c$ We are given the condition $\overrightarrow a \times \overrightarrow b = \overrightarrow c$. Using the anti-commutativity of the cross product, $\overrightarrow b \times \overrightarrow a = -(\overrightarrow a \times \overrightarrow b)$. Substituting the given condition, we get $\overrightarrow b \times \overrightarrow a = -\overrightarrow c$. Now, substitute this back into the expression for the scalar triple product: $(\overrightarrow b \times \overrightarrow a) \cdot \overrightarrow c = (-\overrightarrow c) \cdot \overrightarrow c$ Using the property $\overrightarrow V \cdot \overrightarrow V = |\overrightarrow V|^2$: $(-\overrightarrow c) \cdot \overrightarrow c = -(\overrightarrow c \cdot \overrightarrow c) = -|\overrightarrow c|^2$ We need to find the magnitude of $\overrightarrow c$: $\overrightarrow c = \widehat j - \widehat k$ $|\overrightarrow c| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2}$ Therefore, $|\overrightarrow c|^2 = (\sqrt{2})^2 = 2$. So, the scalar triple product is $\overrightarrow b \cdot (\overrightarrow a \times \overrightarrow c) = -2$. The numerator of $l$ is the absolute value of this result: $|\overrightarrow b \cdot (\overrightarrow a \times \overrightarrow c)| = |-2| = 2$ Note that the condition $\overrightarrow a \cdot \overrightarrow b = 1$ was not needed for this part of the calculation.

Step 4: Calculate 'l'

Now we have the numerator and the denominator for $l$: $l = \frac{|\overrightarrow b \cdot (\overrightarrow a \times \overrightarrow c)|}{|\overrightarrow a \times \overrightarrow c|} = \frac{2}{\sqrt{6}}$

Step 5: Calculate $3l^2$

First, we square $l$: $l^2 = \left(\frac{2}{\sqrt{6}}\right)^2 = \frac{4}{6} = \frac{2}{3}$ Finally, we compute $3l^2$: $3l^2 = 3 \times \frac{2}{3} = 2$

Common Mistakes & Tips

• Scalar Triple Product Manipulation: Recognizing that $\overrightarrow b \cdot (\overrightarrow a \times \overrightarrow c)$ can be rewritten as $(\overrightarrow b \times \overrightarrow a) \cdot \overrightarrow c$ is key.
• Anti-Commutativity: Forgetting the negative sign in $\overrightarrow b \times \overrightarrow a = -(\overrightarrow a \times \overrightarrow b)$ is a common error.
• Absolute Value: Remember that length is always non-negative, so the absolute value is necessary for the numerator of the projection formula.

Summary

The problem required calculating the length of a projection vector. By applying the formula for the projection length and strategically using the properties of the scalar triple product and the cross product, we simplified the expression. The key insight was using the given relation $\overrightarrow a \times \overrightarrow b = \overrightarrow c$ to transform the scalar triple product term into a form involving only $\overrightarrow c$, which allowed for straightforward calculation of its magnitude. The value of $3l^2$ was found to be 2.

The final answer is \boxed{2}.