Q.Find the direction cosines of the vector $\hat{i} + 2\hat{j} + 3\hat{k}$.

We are given three points in space, so the triangle is not necessarily in a plane we can draw easily, but the side lengths are all we need. The cosine of an angle in any triangle depends only on the three side lengths via the law of cosines. So the plan is: compute the distances $AB, BC, CA$, then write $\cos A$ and $\cos B$ in terms of those sides, take the absolute value of their ratio, and simplify.

1. Compute the side lengths. Use the distance formula in 3D:

2. Apply the law of cosines. For angle $A$ (at vertex $A$), the sides meeting at $A$ are $AB$ and $AC$, and the opposite side is $BC$.

Simplify $\sqrt{126} = \sqrt{9 \cdot 14} = 3\sqrt{14}$, so

For angle $B$ (at vertex $B$), the sides meeting at $B$ are $BA$ and $BC$, and the opposite side is $AC$.

$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{6 + 9 - 21}{2 \cdot \sqrt{6} \cdot 3} = \frac{-6}{6\sqrt{6}} = -\frac{1}{\sqrt{6}}.$ …

The problem asks us to evaluate a trigonometric expression involving $\sin \theta$ and $\cos \theta$ for an angle $\theta$ in a right-angled triangle. The key idea is to first determine all the side lengths of the triangle. Once we have all three sides, we can define the sine and cosine ratios for the given angle $\theta$ and then substitute these values into the expression.

In a right-angled triangle, the trigonometric ratios are defined as follows for an angle $\theta$:

• $\sin \theta = \frac{\text{Length of the side opposite to } \theta}{\text{Length of the hypotenuse}}$
• $\cos \theta = \frac{\text{Length of the side adjacent to } \theta}{\text{Length of the hypotenuse}}$

We will use the Pythagorean theorem to find the unknown side length, and then apply these definitions.

1. Find the length of the missing side AC. We are given a right-angled triangle ACB, with the right angle at C. The hypotenuse is AB = 29 units. One leg is BC = 21 units. Let the other leg be AC. According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides:

Substitute the given values:

Calculate the squares:

Now, solve for $AC^2$:

Take the square root to find AC:

2. Determine $\sin \theta$ and $\cos \theta$.

   The angle $\theta$ is $\angle ABC$.

   • The hypotenuse (the side opposite the right angle C) is AB = 29 units.
   • The side opposite to $\angle ABC$ is AC = 20 units.
   • The side adjacent to $\angle ABC$ is BC = 21 units.

   Now, we can write the trigonometric ratios for $\theta$:

   $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{20}{29}$ …