SATHEE: 3D Geometry 1 Question 1

3D Geometry 1 Question 1

The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^{2} = m^{2} + n^{2}$ , is

(a) $\frac{π}{3}$

(b) $\frac{π}{4}$

(d) $\frac{π}{2}$

(2014 Main)

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Answer:

Correct Answer: (a)

Solution:

Steps to Solve:

Relate Direction Cosines:

From $l + m + n = 0$ , express one cosine in terms of others: $l = - m - n$

Substitute into $l^{2} + m^{2} + n^{2} = 1$ to get:

$4 (m^{2} + n^{2}) = 1$

Therefore: $m^{2} + n^{2} = \frac{1}{4}$

Find the Dot Product:

Since both lines are unit vectors, regardless of specific direction cosines:

$(l_{1}, m_{1}, n_{1}) \cdot (l_{2}, m_{2}, n_{2}) = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Find the Angle:

Using the inverse cosine function:

$θ = \cos^{- 1} (\frac{1}{2}) = \frac{π}{3}$

Check for Extraneous Solutions:

The solution $θ = \frac{5 π}{3}$ (cos = -\frac{1}{2}) leads to a contradiction with the given equations.

Therefore, the angle between the lines is:

$θ = \frac{π}{3}$

Method 2

We know that, angle between two lines is

$\cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$

$l + m + n = 0$

$\Rightarrow l = - (m + n) \Rightarrow (m + n)^{2} = l^{2}$

$\Rightarrow m^{2} + n^{2} + 2 m n = m^{2} + n^{2} [∵ l^{2} = m^{2} + n^{2}, given]$

$2 m n = 0$

$When m = 0$

$\Rightarrow l = - n$

Hence, $(l, m, n)$ is $(1, 0, - 1)$ .

When $n = 0$ , then $l = - m$

Hence, $(l, m, n)$ is $(1, 0, - 1)$ .

$∴ \cos θ = \frac{1 + 0 + 0}{\sqrt{2} \times \sqrt{2}} = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$

3D Geometry 1 Question 1

Answer:

Steps to Solve:

Method 2

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3D Geometry 1 Question 1

Answer:

Steps to Solve:

Method 2

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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