Consider the following for the next two (02) items that follow : Let a vector \(\vec{a}=4 \hat{i}-8 \hat{j}+\hat{k}\) make angles α, β, γ with the positive directions of x, y, z axes respectively.
What is cos2β + cos2γ equal to ?
Understanding Vector Angles and Direction Cosines
The problem asks us to find the value of a trigonometric expression involving the angles a given vector makes with the coordinate axes. These angles are related to the direction cosines of the vector.
Let the given vector be \(\vec{a} = 4 \hat{i} - 8 \hat{j} + \hat{k}\). The angles it makes with the positive x, y, and z axes are \(\alpha\), \(\beta\), and \(\gamma\) respectively.
The direction cosines of a vector \(\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}\) are given by:
- \(\cos \alpha = \frac{a_x}{|\vec{a}|}\)
- \(\cos \beta = \frac{a_y}{|\vec{a}|}\)
- \(\cos \gamma = \frac{a_z}{|\vec{a}|}\)
where \(|\vec{a}|\) is the magnitude of the vector.
Calculating Vector Magnitude and Direction Cosines
First, let's calculate the magnitude of the vector \(\vec{a}\):
\(|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}\)
For the given vector \(\vec{a} = 4 \hat{i} - 8 \hat{j} + \hat{k}\), we have \(a_x = 4\), \(a_y = -8\), and \(a_z = 1\).
So, the magnitude is:
\(|\vec{a}| = \sqrt{4^2 + (-8)^2 + 1^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9\)
Now, we can find the direction cosines:
- \(\cos \alpha = \frac{4}{9}\)
- \(\cos \beta = \frac{-8}{9}\)
- \(\cos \gamma = \frac{1}{9}\)
Using Direction Cosines to find cos2β + cos2γ
We need to find the value of \(\cos 2\beta + \cos 2\gamma\). We can use the double angle identity for cosine: \(\cos 2\theta = 2\cos^2 \theta - 1\).
Applying this identity for \(\beta\) and \(\gamma\):
- \(\cos 2\beta = 2\cos^2 \beta - 1\)
- \(\cos 2\gamma = 2\cos^2 \gamma - 1\)
Substitute the values of \(\cos \beta\) and \(\cos \gamma\):
- \(\cos 2\beta = 2\left(\frac{-8}{9}\right)^2 - 1 = 2\left(\frac{64}{81}\right) - 1 = \frac{128}{81} - 1 = \frac{128 - 81}{81} = \frac{47}{81}\)
- \(\cos 2\gamma = 2\left(\frac{1}{9}\right)^2 - 1 = 2\left(\frac{1}{81}\right) - 1 = \frac{2}{81} - 1 = \frac{2 - 81}{81} = \frac{-79}{81}\)
Now, sum these values:
\(\cos 2\beta + \cos 2\gamma = \frac{47}{81} + \frac{-79}{81} = \frac{47 - 79}{81} = \frac{-32}{81}\)
Alternative Method using Direction Cosine Identity
We know the identity relating the squares of direction cosines: \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\).
There is also a related identity for the double angles: \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1\).
From this identity, we can write: \(\cos 2\beta + \cos 2\gamma = -1 - \cos 2\alpha\).
First, calculate \(\cos 2\alpha\):
\(\cos 2\alpha = 2\cos^2 \alpha - 1 = 2\left(\frac{4}{9}\right)^2 - 1 = 2\left(\frac{16}{81}\right) - 1 = \frac{32}{81} - 1 = \frac{32 - 81}{81} = \frac{-49}{81}\)
Now, substitute this into the identity:
\(\cos 2\beta + \cos 2\gamma = -1 - \left(\frac{-49}{81}\right) = -1 + \frac{49}{81} = \frac{-81 + 49}{81} = \frac{-32}{81}\)
Both methods yield the same result.
The value of \(\cos 2\beta + \cos 2\gamma\) is \(\frac{-32}{81}\).
Revision Table: Vector Direction Cosines
| Concept | Description | Formula |
|---|---|---|
| Vector Components | The scalar projections of a vector onto the coordinate axes. | For \(\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}\), components are \(a_x, a_y, a_z\). |
| Vector Magnitude | The length or norm of the vector. | \(|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}\) |
| Direction Cosines | Cosines of the angles a vector makes with the positive coordinate axes. | \(\cos \alpha = \frac{a_x}{|\vec{a}|}\), \(\cos \beta = \frac{a_y}{|\vec{a}|}\), \(\cos \gamma = \frac{a_z}{|\vec{a}|}\) |
| Direction Cosine Identity | Relationship between the squares of direction cosines. | \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\) |
| Double Angle Identity | Relates cosine of double angle to cosine of the angle. | \(\cos 2\theta = 2\cos^2 \theta - 1\) |
| Double Angle Identity (Direction Cosines) | Relationship between double angle cosines of direction angles. | \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1\) |
Additional Information on Vector Angles and Trigonometric Identities
Understanding vector direction cosines and related trigonometric identities is crucial in 3D vector algebra. The angles \(\alpha\), \(\beta\), and \(\gamma\) are often called the direction angles of the vector.
The identity \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\) signifies that the direction cosines are not independent; knowing two allows you to find the third (up to sign).
The identity \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1\) is a direct consequence of the \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\) identity and the double angle formula \(\cos 2\theta = 2\cos^2 \theta - 1\). Substituting \(\cos^2 \theta = (\cos 2\theta + 1)/2\) into the first identity gives:
\(\frac{\cos 2\alpha + 1}{2} + \frac{\cos 2\beta + 1}{2} + \frac{\cos 2\gamma + 1}{2} = 1\)
\(\frac{\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 3}{2} = 1\)
\(\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 3 = 2\)
\(\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 2 - 3 = -1\)
This confirms the identity used in the alternative method. These identities are useful for solving problems involving the orientation of vectors in space without needing to explicitly calculate the angles themselves.
Similar Questions
A point on a line has coordinates (p + 1, p - 3, √2p) where p is any real number. What are the direction cosines of the line?
If l, m, n are the direction cosines of the line x - 1 = 2(y + 3) = 1 - z, then what is l 4+ m 4+ n 4equal to?
A straight line with direction cosines \(\left\langle {0,\;1,\;0} \right\rangle\) is
If a line has direction ratios < a + b, b + c, c + a >, then what is the sum of the squares of its direction cosines?
Consider the following statements :
1. The direction ratios of y-axis can be <0, 4, 0>
2. The direction ratios of a line perpendicular to z-axis can be <5, 6, 0>
Which of the statements given above is/are correct?
The direction ratios of the line perpendicular to the lines with direction ratios < 1, -2, -2 > and < 0, 2, 1 > are
What are the direction cosines of z-axis?
Consider the following statements:
1. A-line in space can have infinitely many direction ratios.
2. It is possible for certain lines that the sum of the squares of direction cosines can be equal to the sum of its direction cosines.
Which of the above statements is/are correct?
If α, β and γ are the angles which the vector \(\overrightarrow {OP} \) (O being the origin) makes with the positive direction of the coordinate axes, then which of the following are correct?
1. cos 2α + cos 2β = sin 2γ
2. sin 2α + sin 2β = cos 2γ
3. sin 2α + sin 2β + sin 2γ = 2
Select the correct answer using the code given below.
What is cosα equal to ?
Important Questions from Direction ratios and Direction cosines
A point on a line has coordinates (p + 1, p - 3, √2p) where p is any real number. What are the direction cosines of the line?
If l, m, n are the direction cosines of the line x - 1 = 2(y + 3) = 1 - z, then what is l 4+ m 4+ n 4equal to?
Which of the following is the direction cosines of z, y and x-axis?
Determine the direction cosines of the unit vector perpendicular to the plane \(\vec{r} \cdot(2 \hat{\imath}-6 \hat{\jmath}-3 \hat{k})+1=0\) passing through the origin?
Determine the direction cosines of the line passing through the two points (-2, 3, -4) and (1, 2, 3)?
