Finding cosα for a Vector: Direction Cosine Explanation

The question asks us to find the value of cosα for a given vector \(\vec{a}=4 \hat{i}-8 \hat{j}+\hat{k}\), where α is the angle the vector makes with the positive x-axis.

The angles α, β, and γ that a vector makes with the positive directions of the x, y, and z axes respectively are called direction angles. The cosines of these angles, cosα, cosβ, and cosγ, are known as the direction cosines of the vector.

For a vector \(\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}\), the direction cosines are given by the following formulas:

• cosα = \(\frac{x}{|\vec{a}|}\)
• cosβ = \(\frac{y}{|\vec{a}|}\)
• cosγ = \(\frac{z}{|\vec{a}|}\)

Here, \(|\vec{a}|\) represents the magnitude of the vector \(\vec{a}\), calculated as \(|\vec{a}| = \sqrt{x^2 + y^2 + z^2}\).

Calculating the Magnitude of the Vector

Given the vector \(\vec{a}=4 \hat{i}-8 \hat{j}+\hat{k}\), the components are \(x=4\), \(y=-8\), and \(z=1\).

Let's calculate the magnitude of \(\vec{a}\):

\[ |\vec{a}| = \sqrt{(4)^2 + (-8)^2 + (1)^2} \]

\[ |\vec{a}| = \sqrt{16 + 64 + 1} \]

\[ |\vec{a}| = \sqrt{81} \]

\[ |\vec{a}| = 9 \]

The magnitude of the vector \(\vec{a}\) is 9.

Finding cosα for the Vector

Now we can find cosα using the formula cosα = \(\frac{x}{|\vec{a}|}\). The x-component of the vector is 4, and the magnitude is 9.

\[ \cos\alpha = \frac{4}{9} \]

So, the value of cosα for the given vector is \(\frac{4}{9}\).

Let's quickly look at the other direction cosines as well, although not required by the question:

• cosβ = \(\frac{y}{|\vec{a}|} = \frac{-8}{9}\)
• cosγ = \(\frac{z}{|\vec{a}|} = \frac{1}{9}\)

A property of direction cosines is that the sum of their squares is always equal to 1: \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\). Let's verify this:

\[ \left(\frac{4}{9}\right)^2 + \left(\frac{-8}{9}\right)^2 + \left(\frac{1}{9}\right)^2 = \frac{16}{81} + \frac{64}{81} + \frac{1}{81} = \frac{16+64+1}{81} = \frac{81}{81} = 1 \]

This confirms our calculations for the direction cosines are correct.

The question specifically asks for cosα, which we found to be \(\frac{4}{9}\).

Revision Table: Vector Direction Cosines

Additional Information: Understanding Vectors and Direction Cosines

A vector in three-dimensional space can be represented as \(\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}\), where x, y, and z are the components of the vector along the x, y, and z axes respectively, and \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) are the unit vectors along these axes.

Direction cosines provide a way to describe the orientation of a vector in space. They are essentially the components of the unit vector in the direction of the given vector. If \(\hat{u}\) is the unit vector in the direction of \(\vec{a}\), then \(\hat{u} = \frac{\vec{a}}{|\vec{a}|} = \frac{x}{|\vec{a}|}\hat{i} + \frac{y}{|\vec{a}|}\hat{j} + \frac{z}{|\vec{a}|}\hat{k}\).

Thus, the components of the unit vector are precisely the direction cosines: cosα = \(\frac{x}{|\vec{a}|}\), cosβ = \(\frac{y}{|\vec{a}|}\), and cosγ = \(\frac{z}{|\vec{a}|}\). This also explains why the sum of the squares of direction cosines is 1, as it is the magnitude squared of the unit vector (\(|\hat{u}|^2 = (\cos\alpha)^2 + (\cos\beta)^2 + (\cos\gamma)^2 = 1^2 = 1\)).

Direction cosines are useful in various applications, including physics and engineering, for describing the direction of forces, velocities, and other vector quantities.