Question

$\alpha ,\beta ,\gamma$ be the angles made by the vectors $3\hat i - 6\hat j + 2\hat k$ with the positive direction of coordinates. Then find $\cos \alpha$, $\cos \beta$ and $\cos \gamma$.

Answer 1

We can find the magnitude of the given vector by taking the square root of the sum of the squares of the components of the vector. Then we can find the required cosines by finding the direction cosine. The direction cosines are given by dividing the respective vector component with the magnitude of the vector.

Complete step by step solution:
We have the vector $3\hat i - 6\hat j + 2\hat k$
Let $r = 3\hat i - 6\hat j + 2\hat k$ .
Now we can find the magnitude of the vector.
We know that magnitude of vector $r = a\hat i + b\hat j + c\hat k$ is given by,
$\left| r \right| = \sqrt {{a^2} + {b^2} + {c^2}}$
On substituting the values, we get,
$\left| r \right| = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( 2 \right)}^2}}$
On taking the squares, we get,
$\left| r \right| = \sqrt {9 + 36 + 4}$
On adding the terms inside the radicle, we get,
$\left| r \right| = \sqrt {49}$
We know that ${7^2} = 49$ . So, we can write,
$\left| r \right| = 7$
Now we have the magnitude of the vector as 7.
We know that $\cos \alpha$ , $\cos \beta$ and $\cos \gamma$ represents the direction cosines of the given vector
We know that equations of the direction cosines are given by,
$\cos \alpha = \dfrac{{{r_x}}}{{\left| r \right|}}$
$\cos \beta = \dfrac{{{r_y}}}{{\left| r \right|}}$
$\cos \gamma = \dfrac{{{r_z}}}{{\left| r \right|}}$
On substituting the values, we get,
$\Rightarrow \cos \alpha = \dfrac{3}{7}$
$\Rightarrow \cos \beta = \dfrac{{ - 6}}{7}$
$\Rightarrow \cos \gamma = \dfrac{2}{7}$

So the required values of the $\cos \alpha$ , $\cos \beta$ and $\cos \gamma$ are $\dfrac{3}{7}$ , $\dfrac{{ - 6}}{7}$ and $\dfrac{2}{7}$

Note:
Alternate solution to this problem is given by,
Let $r = 3\hat i - 6\hat j + 2\hat k$ .
Now we can find the magnitude of the vector.
We know that magnitude of vector $r = a\hat i + b\hat j + c\hat k$ is given by,
$\left| r \right| = \sqrt {{a^2} + {b^2} + {c^2}}$
On substituting the values, we get,
$\left| r \right| = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( 2 \right)}^2}}$
On taking the squares, we get,
$\left| r \right| = \sqrt {9 + 36 + 4}$
$\left| r \right| = \sqrt {49}$
$\left| r \right| = 7$
Now the unit vector is given by,
$\Rightarrow \hat r = \dfrac{r}{{\left| r \right|}}$
$\Rightarrow \hat r = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{7}$
$\Rightarrow \hat r = \dfrac{3}{7}\hat i + \dfrac{{ - 6}}{7}\hat j + \dfrac{2}{7}\hat k$
We know that the components of the unit vectors are the direction cosines of the vector in that direction. So, we can write,
$\Rightarrow \cos \alpha = \dfrac{3}{7}$
$\Rightarrow \cos \beta = \dfrac{{ - 6}}{7}$
$\Rightarrow \cos \gamma = \dfrac{2}{7}$
So the required values of the $\cos \alpha$ , $\cos \beta$ and $\cos \gamma$ are $\dfrac{3}{7}$ , $\dfrac{{ - 6}}{7}$ and $\dfrac{2}{7}$
We must note that the direction cosines are the cosines of the angle that the vector makes with positive axes. The direction ratios are any multiple of the direction cosines. The sum of the squares of the direction cosines will be always equal to 1.