Question

Let $\vec{u}$, $\vec{v}$ and $\vec{w}$ be the vectors such that $|\vec{u}| = 1$; $|\vec{v}| = 2$; $|\vec{w}| = 3$. If the projection of $\vec{v}$ along $\vec{u}$ is equal to that of $\vec{w}$ along $\vec{u}$ and $\vec{v}$, $\vec{w}$ are perpendicular to each other, then $|\vec{u} - \vec{v} + \vec{w}|$ is equal to}

• A. 2
• B. $\sqrt{7}$
• C. $\sqrt{14}$
• D. 14

Hint:

When finding the magnitude of a sum or difference of vectors, square the expression first. This converts it into a dot product, allowing you to use magnitudes and dot product conditions more easily. Remember that $2(\vec{u} \cdot \vec{w}) - 2(\vec{u} \cdot \vec{v})$ simplifies to $0$ if $\vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w}$.

Answer 1

The Correct Option is A

Step 1: Use the condition about projections. The projection of $\vec{v}$ along $\vec{u}$ is given by $\frac{\vec{v} \cdot \vec{u{|\vec{u}|}$. The projection of $\vec{w}$ along $\vec{u}$ is given by $\frac{\vec{w} \cdot \vec{u{|\vec{u}|}$. Given that these projections are equal: \[ \frac{\vec{v} \cdot \vec{u{|\vec{u}|} = \frac{\vec{w} \cdot \vec{u{|\vec{u}|} \] Since $|\vec{u}| = 1 \neq 0$, we can multiply both sides by $|\vec{u}|$: \[ \vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u} \] This implies $\vec{u} \cdot \vec{v} - \vec{u} \cdot \vec{w} = 0$, or $\vec{u} \cdot (\vec{v} - \vec{w}) = 0$. (Equation 1)
Step 2: Use the condition that $\vec{v$ and $\vec{w}$ are perpendicular.} If two vectors are perpendicular, their dot product is zero. \[ \vec{v} \cdot \vec{w} = 0 \quad (\text{Equation 2}) \]
Step 3: Calculate the square of the required magnitude. We need to find $|\vec{u} - \vec{v} + \vec{w}|$. Let's first calculate its square: \[ |\vec{u} - \vec{v} + \vec{w}|^2 = (\vec{u} - \vec{v} + \vec{w}) \cdot (\vec{u} - \vec{v} + \vec{w}) \] Expanding the dot product using the property $(\vec{A} + \vec{B} + \vec{C}) \cdot (\vec{A} + \vec{B} + \vec{C}) = |\vec{A}|^2 + |\vec{B}|^2 + |\vec{C}|^2 + 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A})$ (adjusted for signs): \[ |\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u} \cdot \vec{v}) + 2(\vec{u} \cdot \vec{w}) - 2(\vec{v} \cdot \vec{w}) \]
Step 4: Substitute the given values and derived conditions. Given magnitudes: $|\vec{u}| = 1$, $|\vec{v}| = 2$, $|\vec{w}| = 3$. From Equation 1: $\vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w}$. This means $-2(\vec{u} \cdot \vec{v}) + 2(\vec{u} \cdot \vec{w}) = 0$. From Equation 2: $\vec{v} \cdot \vec{w} = 0$. Substitute these into the expanded expression: \[ |\vec{u} - \vec{v} + \vec{w}|^2 = (1)^2 + (2)^2 + (3)^2 + 0 - 2(0) \] \[ |\vec{u} - \vec{v} + \vec{w}|^2 = 1 + 4 + 9 \] \[ |\vec{u} - \vec{v} + \vec{w}|^2 = 14 \]
Step 5: Find the magnitude. Take the square root of both sides: \[ |\vec{u} - \vec{v} + \vec{w}| = \sqrt{14} \] Thus, $|\vec{u} - \vec{v} + \vec{w}|$ is equal to $\sqrt{14}$.