Q.(b) A person standing at $O(0, 0, 0)$ is watching an aeroplane which is at the coordinate point $A(4, 0, 3)$. At the same time he saw a bird at the coordinate point $B(0, 0, 1)$. Find the angles which $\overrightarrow{BA}$ makes with the $x$, $y$ and $z$ axes.

Concept: Direction Vectors — the vector from $B$ to $A$ is $\overrightarrow{BA} = A - B$, and its direction cosines give the angles with the axes.

Step 1: Find $\overrightarrow{BA}$

Step 2: Compute its magnitude

Step 3: Direction cosines and angles

Thus: …

The problem asks for the angles a vector makes with the coordinate axes. That is a question about direction cosines — the cosines of the angles between the vector and each positive axis. If you know the components of the vector, you can find these directly.

First, what is $\overrightarrow{BA}$? The notation $\overrightarrow{BA}$ means the vector that starts at $B$ and ends at $A$. So it is $A - B$:

Now, the angle $\alpha$ with the $x$-axis, $\beta$ with the $y$-axis, and $\gamma$ with the $z$-axis are given by:

So we need the magnitude of $\overrightarrow{BA}$:

Now compute each direction cosine:

1. Angle with the $x$-axis

   $\cos\alpha = \dfrac{4}{2\sqrt{5}} = \dfrac{2}{\sqrt{5}}$.

   So $\alpha = \cos^{-1}\!\left(\dfrac{2}{\sqrt{5}}\right)$.

2. Angle with the $y$-axis

   $\cos\beta = \dfrac{0}{2\sqrt{5}} = 0$.

   So $\beta = \cos^{-1}(0) = 90^\circ$.

3. Angle with the $z$-axis

   $\cos\gamma = \dfrac{2}{2\sqrt{5}} = \dfrac{1}{\sqrt{5}}$. …

Method: Angles a vector makes with the coordinate axes

Use this to find the angles (equivalently, the direction cosines) that a vector joining two points makes with the $x, y, z$ axes.

Steps

Step 1: Build the direction vector.

For the vector from $B$ to $A$, compute $\vec{BA}=A-B$ (tip minus tail). Get the order right — $\vec{BA}$, not $\vec{AB}$.

Step 2: Find its magnitude.

Step 3: Form the direction cosines, then the angles. …

Common Mistakes

Mistake 1: Forgetting to divide the components by the magnitude.

Why it's wrong: the components $(4,0,2)$ are direction RATIOS, not cosines; the direction cosines require dividing by $|\vec{BA}|=2\sqrt5$. Correct approach: $\cos\alpha=\frac{4}{2\sqrt5}$, and so on.

Mistake 2: Building $\vec{AB}$ instead of $\vec{BA}$. …