Q.Find the unit vector in the direction of the sum of the vectors, $\vec{a}=2\hat{i}+2\hat{j}-5\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+3\hat{k}$.

Concept and Intuition

The line's direction is the difference of the two position vectors; normalize it to length $1$.

Step-by-Step Solution

1. Direction $=(13-10,\,9-8,\,4-6)=(3,1,-2)$.
2. $|(3,1,-2)|=\sqrt{9+1+4}=\sqrt{14}$.
3. Unit vector $=\dfrac{1}{\sqrt{14}}(3\hat i+\hat j-2\hat k)$.

Common Mistakes …

With $\vec{OQ}=2\hat i+\hat j+2\hat k$, $|\vec{OQ}|=3$, and $\vec{OP}=2\hat i-2\hat j-\hat k$ (so $|\vec{OP}|=3$), the two sides of $\angle POQ$ are equal in length.

The internal bisector meets $PQ$ at $R$ with $PR:RQ=|OP|:|OQ|=1:1$, i.e. $R$ is the midpoint of $PQ$. …

$\vec{r}=(2-3t)(\hat{i}+\hat{j})+(6t+1)(\hat{j}+\hat{k})+(12t-11)(\hat{k}+\hat{i})$.

$\hat{i}$: $(2-3t)+(12t-11)=9t-9$.

$\hat{j}$: $(2-3t)+(6t+1)=3t+3$.

$\hat{k}$: $(6t+1)+(12t-11)=18t-10$. …

Direction $=(3-1,\,2-(-1),\,8-2)=(2,3,6)$ with magnitude $\sqrt{4+9+36}=7$. …

The direction ratios of the line joining $(2,3,4)$ and $(-1,4,-3)$ are

$(-1-2,\ 4-3,\ -3-4)=(-3,\ 1,\ -7).$ …

The direction vector of $\vec{r}=(\hat{i}-11\hat{j}+101\hat{k})+\lambda(3\hat{i}-5\hat{j}+2\hat{k})$ is $\vec{d}=3\hat{i}-5\hat{j}+2\hat{k}$. …

For angles $\dfrac{\pi}{3},\dfrac{\pi}{4},\gamma$ made with the axes,

Since $\cos^2\dfrac{\pi}{3}=\dfrac{1}{4}$ and $\cos^2\dfrac{\pi}{4}=\dfrac{1}{2}$, …

Concept and Intuition

Translating a point a distance $1$ in the direction of a line means adding one unit vector along that line's direction.

Step-by-Step Solution

1. Direction of $y=2x$ is $(1,2)$; unit vector $=\tfrac{1}{\sqrt5}(1,2)$.
2. $Q=(1,1)+\tfrac{1}{\sqrt5}(1,2)=\left(1+\tfrac{1}{\sqrt5},\,1+\tfrac{2}{\sqrt5}\right)$. …

A line equally inclined to the coordinate axes has direction ratios $(1,1,1)$. Through $(1,-2,3)$: …