Q.If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} - 2\hat{j} + \hat{k}$, find a unit vector parallel to the vector $2\vec{a} - \vec{b} + 3\vec{c}$.

Concept and Intuition

"In the plane of $a$ and $b$" means the vector is a linear combination $x a+y b$. Imposing perpendicularity to $a$ gives one linear constraint on $x,y$, typically leaving a 1-parameter family (a line through the origin) — exactly enough freedom to pick out a specific direction, then normalize.

Step-by-Step Solution

1. Write a general vector in the plane: $v=x(1,1,0)+y(1,1,1)=(x+y,\ x+y,\ y)$.
2. Impose $v\cdot a=0$: $(x+y)(1)+(x+y)(1)+y(0)=2(x+y)=0 \Rightarrow x+y=0 \Rightarrow x=-y$.
3. Substitute back: $v=(x+y,\ x+y,\ y)=(0,0,y)$.
4. So $v$ is a scalar multiple of $(0,0,1)=\hat k$, and it is already a unit vector when $y=\pm1$; taking $y=1$ gives $\hat k$.

Common Mistakes …

Concept and Intuition

The internal angle bisector between two vectors $\bar a$ and $\bar b$ (from a common vertex) points along $\hat a + \hat b$ (sum of unit vectors), because that sum lies symmetrically between the two directions with equal projections onto each.

Step-by-Step Solution

1. Let $\bar b = -2\bar i - \bar j + 2\bar k$, so $|\bar b| = \sqrt{4+1+4}=3$, giving $\hat b = \left(-\tfrac23,-\tfrac13,\tfrac23\right)$.
2. Let $\hat a = (x,y,z)$ be the unit vector along $\bar a$. The bisector direction $\hat a+\hat b$ must be parallel to $\bar i-7\bar j+2\bar k$: $\left(x-\tfrac23,\ y-\tfrac13,\ z+\tfrac23\right) = k(1,-7,2)$
3. So $x=k+\tfrac23,\ y=\tfrac13-7k,\ z=2k-\tfrac23$.
4. Impose $x^2+y^2+z^2=1$: expanding gives $54k^2-6k+1=1 \Rightarrow 54k^2-6k=0 \Rightarrow 6k(9k-1)=0$, so $k=0$ or $k=\tfrac19$. …

Concept and Intuition

$\bar u=\bar a-(\bar a.\bar b)\bar b$ is the standard Gram–Schmidt construction: subtracting off the projection of $\bar a$ along $\bar b$ leaves the component of $\bar a$ perpendicular to $\bar b$, which still lies in the plane containing $\bar a$ and $\bar b$. The cross product $\bar v=\bar a\times\bar b$ is always perpendicular to both $\bar a$ and $\bar b$, hence perpendicular to their entire plane — so $\bar u$ and $\bar v$ are automatically perpendicular.

Step-by-Step Solution

1. Since $|\bar b|=1$: $\bar u.\bar b=\bar a.\bar b-(\bar a.\bar b)(\bar b.\bar b)=\bar a.\bar b-\bar a.\bar b=0$. So $\bar u\perp\bar b$.
2. $\bar u$ is a linear combination of $\bar a$ and $\bar b$, so it lies in their plane; $\bar v=\bar a\times\bar b$ is normal to that plane, so $\bar u.\bar v=0$.
3. Compute magnitudes directly with $|\bar a|=|\bar b|=1$: $|\bar u|^2=|\bar a|^2-2(\bar a.\bar b)^2+(\bar a.\bar b)^2=1-(\bar a.\bar b)^2$. …