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JEE Main 2019
3D Geometry
3D Geometry
Medium

Question

A line makes the same angle θ\theta , with each of the xx and zz axis. If the angle β\beta \,, which it makes with y-axis, is such that sin2β=3sin2θ,\,{\sin ^2}\beta = 3{\sin ^2}\theta , then cos2θ{\cos ^2}\theta equals :

Options

Solution

Key Concepts and Formulas

  • Direction Cosines of a Line: For a line in 3D space, if it makes angles α\alpha, β\beta, and γ\gamma with the positive xx, yy, and zz-axes respectively, then cosα\cos\alpha, cosβ\cos\beta, and cosγ\cos\gamma are called its direction cosines. They are often denoted as ll, mm, and nn.
  • Fundamental Identity of Direction Cosines: The sum of the squares of the direction cosines of any line is always equal to 1. cos2α+cos2β+cos2γ=1{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1
  • Pythagorean Trigonometric Identity: For any angle xx, the following identity holds: sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1 This can be rearranged as cos2x=1sin2x{\cos ^2}x = 1 - {\sin ^2}x or sin2x=1cos2x{\sin ^2}x = 1 - {\cos ^2}x.

Step-by-Step Solution

Step 1: Identify the angles the line makes with the coordinate axes. The problem states that the line makes an angle θ\theta with the xx-axis, an angle β\beta with the yy-axis, and an angle θ\theta with the zz-axis.

  • Why this step: This is the initial interpretation of the given information, mapping the problem's description to the standard notation for direction cosines. Therefore, we have:
  • Angle with xx-axis, α=θ\alpha = \theta
  • Angle with yy-axis, β=β\beta = \beta
  • Angle with zz-axis, γ=θ\gamma = \theta

Step 2: Apply the fundamental identity of direction cosines. Substitute the identified angles into the fundamental identity cos2α+cos2β+cos2γ=1{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1.

  • Why this step: This identity is the core geometric principle that connects the angles a line makes with the coordinate axes. It forms the primary equation for solving the problem. Substituting the values, we get: cos2θ+cos2β+cos2θ=1{\cos ^2}\theta + {\cos ^2}\beta + {\cos ^2}\theta = 1

Step 3: Simplify the equation. Combine the like terms involving cos2θ{\cos ^2}\theta.

  • Why this step: Simplifying the equation makes it easier to manipulate algebraically. 2cos2θ+cos2β=12{\cos ^2}\theta + {\cos ^2}\beta = 1

Step 4: Use the Pythagorean trigonometric identity to relate cos2β\cos^2\beta and sin2β\sin^2\beta. The problem provides a condition involving sin2β{\sin ^2}\beta. To use this condition, we need to express cos2β{\cos ^2}\beta in terms of sin2β{\sin ^2}\beta using the identity sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1.

  • Why this step: This transformation is crucial for connecting our derived equation with the specific condition given in the problem statement, which involves sin2β{\sin ^2}\beta. From the identity, we know cos2β=1sin2β{\cos ^2}\beta = 1 - {\sin ^2}\beta. Substitute this into the equation from Step 3: 2cos2θ+(1sin2β)=12{\cos ^2}\theta + (1 - {\sin ^2}\beta) = 1 Rearrange the terms to isolate sin2β{\sin ^2}\beta: 2cos2θsin2β=112{\cos ^2}\theta - {\sin ^2}\beta = 1 - 1 2cos2θ=sin2β2{\cos ^2}\theta = {\sin ^2}\beta

Step 5: Substitute the given condition into the equation. The problem states the relationship: sin2β=3sin2θ{\sin ^2}\beta = 3{\sin ^2}\theta.

  • Why this step: This is a critical substitution that eliminates the angle β\beta from our equation, leaving an equation solely in terms of θ\theta, which is what we need to solve for. Substitute this into the equation from Step 4: 2cos2θ=3sin2θ2{\cos ^2}\theta = 3{\sin ^2}\theta

Step 6: Convert sin2θ\sin^2\theta to cos2θ\cos^2\theta and solve. To find the value of cos2θ{\cos ^2}\theta, we need to express the entire equation in terms of cos2θ{\cos ^2}\theta. We use the Pythagorean identity again: sin2θ=1cos2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta.

  • Why this step: This allows us to have a single variable (cos2θ{\cos ^2}\theta) in the equation, making it solvable algebraically. Substitute this into the equation from Step 5: 2cos2θ=3(1cos2θ)2{\cos ^2}\theta = 3(1 - {\cos ^2}\theta)

Step 7: Expand and solve for cos2θ\cos^2\theta. Distribute the 3 on the right side and collect all terms involving cos2θ{\cos ^2}\theta.

  • Why this step: These are standard algebraic steps to isolate the desired variable and find its value. 2cos2θ=33cos2θ2{\cos ^2}\theta = 3 - 3{\cos ^2}\theta Add 3cos2θ3{\cos ^2}\theta to both sides: 2cos2θ+3cos2θ=32{\cos ^2}\theta + 3{\cos ^2}\theta = 3 5cos2θ=35{\cos ^2}\theta = 3 Divide by 5: cos2θ=35{\cos ^2}\theta = {3 \over 5}

Common Mistakes & Tips

  • Forgetting the fundamental identity: The relation cos2α+cos2β+cos2γ=1{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 is the cornerstone of 3D geometry problems involving angles a line makes with the axes. Always start by writing it down.
  • Errors in trigonometric identities: Be careful when converting between sine and cosine squares (e.g., 1cos2x=sin2x1 - {\cos ^2}x = {\sin ^2}x).
  • Algebraic blunders: Pay close attention to distribution, combining like terms, and rearranging equations. Simple calculation errors can lead to an incorrect final answer.
  • Reading the question: Ensure you are solving for the quantity asked (here, cos2θ{\cos ^2}\theta, not cosθ\cos\theta or θ\theta).

Summary

This problem effectively combines the geometric concept of direction cosines with fundamental trigonometric identities. By first applying the direction cosine identity, we established a relationship between the angles. Then, using the Pythagorean identity and the given condition, we systematically transformed the equation to express it solely in terms of cos2θ{\cos ^2}\theta. Finally, algebraic manipulation yielded the required value.

The final answer is 35\boxed{{3 \over 5}}, which corresponds to option (C).

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