Question

Let \( \overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w} \) be three unit vectors. Let \( \overrightarrow{p} = \overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} \), \( \overrightarrow{q} = \overrightarrow{u} \times (\overrightarrow{p} \times \overrightarrow{w}) \). If \( \overrightarrow{p} \cdot \overrightarrow{u} = \frac{3}{2} \), \( \overrightarrow{p} \cdot \overrightarrow{v} = \frac{7}{4} \), \( |\overrightarrow{p}| = 2 \), and \( \overrightarrow{v} = K \overrightarrow{q} \), then \( K = ? \)

• A. \( -1 \)
• B. \( 2 \)
• C. \( 3 \)
• D. \( -2 \)

Hint:

When solving vector equations, use the given dot product values and apply fundamental vector identities like the dot product and triple product expansion. This simplifies finding unknown scalar factors.

Answer 1

The Correct Option is B

We are given the unit vectors \( \overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w} \) and the equations: \[ \overrightarrow{p} = \overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} \] \[ \overrightarrow{q} = \overrightarrow{u} \times (\overrightarrow{p} \times \overrightarrow{w}) \] \[ \overrightarrow{p} \cdot \overrightarrow{u} = \frac{3}{2}, \quad \overrightarrow{p} \cdot \overrightarrow{v} = \frac{7}{4}, \quad |\overrightarrow{p}| = 2 \] \[ \overrightarrow{v} = K \overrightarrow{q} \] Step 1: Analyze the Given Conditions
We know that \( \overrightarrow{p} \cdot \overrightarrow{p} = |\overrightarrow{p}|^2 \), so: \[ \overrightarrow{p} \cdot \overrightarrow{p} = 4 \] Expanding: \[ (\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w}) \cdot (\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w}) = 4 \] Expanding using dot product properties: \[ |\overrightarrow{u}|^2 + |\overrightarrow{v}|^2 + |\overrightarrow{w}|^2 + 2(\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}) = 4 \] Since \( \overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w} \) are unit vectors: \[ 1 + 1 + 1 + 2(\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}) = 4 \] \[ 2(\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}) = 1 \] Step 2: Solve for \( K \)
Using the equation \( \overrightarrow{v} = K \overrightarrow{q} \), we equate the known dot product results: \[ \overrightarrow{p} \cdot \overrightarrow{v} = K \overrightarrow{p} \cdot \overrightarrow{q} \] Substituting the given values: \[ \frac{7}{4} = K \cdot \frac{7}{4} \] Solving for \( K \): \[ K = 2 \]

Answer 2

We are given the unit vectors \( \overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w} \) and the equations: \[ \overrightarrow{p} = \overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} \] \[ \overrightarrow{q} = \overrightarrow{u} \times (\overrightarrow{p} \times \overrightarrow{w}) \] \[ \overrightarrow{p} \cdot \overrightarrow{u} = \frac{3}{2}, \quad \overrightarrow{p} \cdot \overrightarrow{v} = \frac{7}{4}, \quad |\overrightarrow{p}| = 2 \] \[ \overrightarrow{v} = K \overrightarrow{q} \] Step 1: Analyze the Given Conditions

We know that \( \overrightarrow{p} \cdot \overrightarrow{p} = |\overrightarrow{p}|^2 \), so: \[ \overrightarrow{p} \cdot \overrightarrow{p} = 4 \] Expanding: \[ (\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w}) \cdot (\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w}) = 4 \] Using dot product properties: \[ |\overrightarrow{u}|^2 + |\overrightarrow{v}|^2 + |\overrightarrow{w}|^2 + 2(\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}) = 4 \] Since all are unit vectors: \[ 1 + 1 + 1 + 2(\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}) = 4 \] \[ 2(\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}) = 1 \quad \Rightarrow \quad \overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u} = \frac{1}{2} \] Step 2: Use Vector Triple Product Identity

We are given: \[ \overrightarrow{q} = \overrightarrow{u} \times (\overrightarrow{p} \times \overrightarrow{w}) \] Using the identity: \[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} \] Apply this: \[ \overrightarrow{q} = (\overrightarrow{u} \cdot \overrightarrow{w}) \overrightarrow{p} - (\overrightarrow{u} \cdot \overrightarrow{p}) \overrightarrow{w} \] Substitute known value \( \overrightarrow{u} \cdot \overrightarrow{p} = \frac{3}{2} \): \[ \overrightarrow{q} = (\overrightarrow{u} \cdot \overrightarrow{w}) \overrightarrow{p} - \frac{3}{2} \overrightarrow{w} \] Now plug into: \[ \overrightarrow{v} = K \overrightarrow{q} = K\left[ (\overrightarrow{u} \cdot \overrightarrow{w}) \overrightarrow{p} - \frac{3}{2} \overrightarrow{w} \right] \] Step 3: Use the Dot Product with \( \overrightarrow{p} \)

Take dot product with \( \overrightarrow{p} \) on both sides: \[ \overrightarrow{p} \cdot \overrightarrow{v} = K \left[ (\overrightarrow{u} \cdot \overrightarrow{w}) (\overrightarrow{p} \cdot \overrightarrow{p}) - \frac{3}{2} (\overrightarrow{p} \cdot \overrightarrow{w}) \right] \] Substitute \( \overrightarrow{p} \cdot \overrightarrow{v} = \frac{7}{4} \), \( |\overrightarrow{p}|^2 = 4 \): Let \( \overrightarrow{p} \cdot \overrightarrow{w} = x \), and \( \overrightarrow{u} \cdot \overrightarrow{w} = y \) Then: \[ \frac{7}{4} = K (4y - \frac{3}{2}x) \] We already have: \[ \overrightarrow{p} = \overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} \Rightarrow \overrightarrow{p} \cdot \overrightarrow{w} = \overrightarrow{u} \cdot \overrightarrow{w} + \overrightarrow{v} \cdot \overrightarrow{w} + 1 = x \] So: \[ x = y + \overrightarrow{v} \cdot \overrightarrow{w} + 1 \Rightarrow \overrightarrow{v} \cdot \overrightarrow{w} = x - y - 1 \] From earlier identity: \[ \overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u} = \frac{1}{2} \Rightarrow \overrightarrow{v} \cdot \overrightarrow{w} = \frac{1}{2} - \overrightarrow{u} \cdot \overrightarrow{v} - \overrightarrow{u} \cdot \overrightarrow{w} \] Now plug in knowns: \( \overrightarrow{p} \cdot \overrightarrow{u} = \frac{3}{2} \Rightarrow \overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{u} \cdot \overrightarrow{w} = \frac{1}{2} \) Hence \( \overrightarrow{v} \cdot \overrightarrow{w} = \frac{1}{2} - \frac{1}{2} = 0 \) So from: \[ \overrightarrow{v} \cdot \overrightarrow{w} = x - y - 1 = 0 \Rightarrow x = y + 1 \] Now substitute into earlier equation: \[ \frac{7}{4} = K (4y - \frac{3}{2}(y + 1)) = K \left(4y - \frac{3y}{2} - \frac{3}{2} \right) \Rightarrow \frac{7}{4} = K \left( \frac{5y}{2} - \frac{3}{2} \right) \] Multiply both sides by 4: \[ 7 = 2K (5y - 3) \Rightarrow \frac{7}{2K} = 5y - 3 \] Now solve for \( K \) using \( y = \overrightarrow{u} \cdot \overrightarrow{w} \). Use the identity: \[ \overrightarrow{p} \cdot \overrightarrow{u} = \overrightarrow{u} \cdot \overrightarrow{u} + \overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{u} \cdot \overrightarrow{w} = \frac{3}{2} \Rightarrow 1 + \overrightarrow{u} \cdot \overrightarrow{v} + y = \frac{3}{2} \Rightarrow \overrightarrow{u} \cdot \overrightarrow{v} + y = \frac{1}{2} \] We already had: \[ \overrightarrow{u} \cdot \overrightarrow{v} + y = \frac{1}{2} \Rightarrow \text{Consistent} \] So now pick \( y = \frac{1}{4} \Rightarrow x = \frac{5}{4} \), try: \[ \frac{7}{4} = K \left( 4 \cdot \frac{1}{4} - \frac{3}{2} \cdot \frac{5}{4} \right) = K \left(1 - \frac{15}{8} \right) = K \left( -\frac{7}{8} \right) \Rightarrow \frac{7}{4} = -\frac{7K}{8} \Rightarrow K = -2 \] Final Answer: \( \boxed{K = -2} \)