Q73 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
Let $\bar{u}, \bar{v}, \bar{w}$ be three vectors such that $|\bar{u}| = 1, |\bar{v}| = 2, |\bar{w}| = 3$. If the projection of $\bar{v}$ along $\bar{u}$ is equal to the projection of $\bar{w}$ along $\bar{u}$ and $\bar{v}, \bar{w}$ are perpendicular to each other, then $|\bar{u} - \bar{v} + \bar{w}| = $...
- A$4$
- B$\sqrt{7}$
- C$2$
- Correct answerD$\sqrt{14}$
Solution
Given $|\vec{u}| = 1$, $|\vec{v}| = 2$, and $|\vec{w}| = 3$.
The projection of $\vec{v}$ along $\vec{u}$ is equal to the projection of $\vec{w}$ along $\vec{u}$.
$\dfrac{\vec{v} \cdot \vec{u}}{|\vec{u}|} = \dfrac{\vec{w} \cdot \vec{u}}{|\vec{u}|}$
$\vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u}$
Since $\vec{v}$ and $\vec{w}$ are perpendicular to each other, their dot product is zero:
$\vec{v} \cdot \vec{w} = 0$
We need to find the value of $|\vec{u} - \vec{v} + \vec{w}|$. Consider its square:
$|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |-\vec{v}|^2 + |\vec{w}|^2 + 2(\vec{u} \cdot (-\vec{v})) + 2((-\vec{v}) \cdot \vec{w}) + 2(\vec{u} \cdot \vec{w})$
$|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u} \cdot \vec{v}) - 2(\vec{v} \cdot \vec{w}) + 2(\vec{u} \cdot \vec{w})$
Substituting $\vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w}$ and $\vec{v} \cdot \vec{w} = 0$ into the expansion:
$|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u} \cdot \vec{w}) - 2(0) + 2(\vec{u} \cdot \vec{w})$
$|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2$
Substituting the given magnitudes:
$|\vec{u} - \vec{v} + \vec{w}|^2 = (1)^2 + (2)^2 + (3)^2$
$|\vec{u} - \vec{v} + \vec{w}|^2 = 1 + 4 + 9 = 14$
Taking the square root on both sides:
$|\vec{u} - \vec{v} + \vec{w}| = \sqrt{14}$
The projection of $\vec{v}$ along $\vec{u}$ is equal to the projection of $\vec{w}$ along $\vec{u}$.
$\dfrac{\vec{v} \cdot \vec{u}}{|\vec{u}|} = \dfrac{\vec{w} \cdot \vec{u}}{|\vec{u}|}$
$\vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u}$
Since $\vec{v}$ and $\vec{w}$ are perpendicular to each other, their dot product is zero:
$\vec{v} \cdot \vec{w} = 0$
We need to find the value of $|\vec{u} - \vec{v} + \vec{w}|$. Consider its square:
$|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |-\vec{v}|^2 + |\vec{w}|^2 + 2(\vec{u} \cdot (-\vec{v})) + 2((-\vec{v}) \cdot \vec{w}) + 2(\vec{u} \cdot \vec{w})$
$|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u} \cdot \vec{v}) - 2(\vec{v} \cdot \vec{w}) + 2(\vec{u} \cdot \vec{w})$
Substituting $\vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w}$ and $\vec{v} \cdot \vec{w} = 0$ into the expansion:
$|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u} \cdot \vec{w}) - 2(0) + 2(\vec{u} \cdot \vec{w})$
$|\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2$
Substituting the given magnitudes:
$|\vec{u} - \vec{v} + \vec{w}|^2 = (1)^2 + (2)^2 + (3)^2$
$|\vec{u} - \vec{v} + \vec{w}|^2 = 1 + 4 + 9 = 14$
Taking the square root on both sides:
$|\vec{u} - \vec{v} + \vec{w}| = \sqrt{14}$
Hence, the correct answer is option: 'D'.