Let $L$ be a straight line passing through the origin, and it makes angles $\alpha, \beta$ and $\gam...

The relationship between the direction cosines of a line is given by:$ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $Using the identity for double angles:$ \cos 2x = 2\cos^2 x - 1 $We can express the sum $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ as follows:$ \begin{align*} \cos 2\alpha + \cos 2\beta + \cos 2\gamma & = (2\cos^2 \alpha - 1) + (2\cos^2 \beta - 1) + (2\cos^2 \gamma - 1) \\ & = 2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) - 3 \\ & = 2 \cdot 1 - 3 \quad \text{(since } \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\text{)} \\ & = 2 - 3 \\ & = -1 \end{align*} $Thus, the value of $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ is $-1$.

Consider the lines $L_1$ and $L_2$ given below:$$ \begin{gathered} L 1: x=2+\lambda, y=3+2 \lambda, ...

The given lines are:$ \begin{aligned} L_1: & \quad x = 2 + \lambda, \quad y = 3 + 2\lambda, \quad z = 4 + 3\lambda \\ L_2: & \quad x = 4 + \mu, \quad y = 4, \quad z = 4 + \mu \end{aligned} $The point $ (2, 3, 4) $ on $ L_1 $ is closest to $ L_2 $. To find the point on $ L_2 $ closest to $ L_1 $, we use the property that the vector connecting the closest points on the lines is perpendicular to both direction vectors.Direction Vectors:$ \mathbf{d}_1 = \langle 1, 2, 3 \rangle $ for $ L_1 $$ \mathbf{d}_2 = \langle 1, 0, 1 \rangle $ for $ L_2 $Parametric Points:A point on $ L_1 $: $ P_1(\lambda) = (2 + \lambda, 3 + 2\lambda, 4 + 3\lambda) $A point on $ L_2 $: $ P_2(\mu) = (4 + \mu, 4, 4 + \mu) $Vector Between Points:$ \begin{aligned} \mathbf{P}_1 - \mathbf{P}_2 &= \langle (2+\lambda) - (4+\mu), (3+2\lambda) - 4, (4+3\lambda) - (4+\mu) \rangle \\ &= \langle -2 + \lambda - \mu, -1 + 2\lambda, 3\lambda - \mu \rangle \end{aligned} $Perpendicular Conditions:$\mathbf{P}_1 - \mathbf{P}_2$ perpendicular to $\mathbf{d}_1$:$ (-2 + \lambda - \mu) \cdot 1 + (-1 + 2\lambda) \cdot 2 + (3\lambda - \mu) \cdot 3 = 0 $Expanding this:$ -2 + \lambda - \mu - 2 + 4\lambda + 9\lambda - 3\mu = 0 $Simplifies to:$ 14\lambda - 4\mu - 4 = 0 $Therefore:$ 7\lambda - 2\mu = 2 \quad \text{(Equation 1)} $$\mathbf{P}_1 - \mathbf{P}_2$ perpendicular to $\mathbf{d}_2$:$ (-2 + \lambda - \mu) \cdot 1 + (-1 + 2\lambda) \cdot 0 + (3\lambda - \mu) \cdot 1 = 0 $Expanding this:$ -2 + \lambda - \mu + 3\lambda - \mu = 0 $Simplifies to:$ 4\lambda - 2\mu = 2 \quad \text{(Equation 2)} $Solving Equations:From Equation 2:$ 2\lambda - \mu = 1 \quad \Rightarrow \quad \mu = 2\lambda - 1 $Substitute into Equation 1:$ 7\lambda - 2(2\lambda - 1) = 2 $Simplifies to:$ 7\lambda - 4\lambda + 2 = 2 $$ 3\lambda = 0 \quad \Rightarrow \quad \lambda = 0 $Substituting $\lambda = 0$ into $\mu = 2\lambda - 1$:$ \mu = -1 $Closest Point on $ L_2 $:Substitute $\mu = -1$ into the equation for $ L_2 $:$ P_2(-1) = (4 + (-1), 4, 4 + (-1)) = (3, 4, 3) $Thus, the point on $ L_2 $ that is closest to $ L_1 $ is $ (3, 4, 3) $.

Three Dimensional Geometry - IAT-IISER Maths PYQs

The given lines are:

%5Cbegin%7Baligned%7D%20L_1%3A%20%26%20%5Cquad%20x%20%3D%202%20%2B%20%5Clambda%2C%20%5Cquad%20y%20%3D%203%20%2B%202%5Clambda%2C%20%5Cquad%20z%20%3D%204%20%2B%203%5Clambda%20%5C%5C%20L_2%3A%20%26%20%5Cquad%20x%20%3D%204%20%2B%20%5Cmu%2C%20%5Cquad%20y%20%3D%204%2C%20%5Cquad%20z%20%3D%204%20%2B%20%5Cmu%20%5Cend%7Baligned%7D

The point

(2%2C%203%2C%204)

L_1

is closest to

L_2

. To find the point on

L_2

closest to

L_1

, we use the property that the vector connecting the closest points on the lines is perpendicular to both direction vectors.Direction Vectors:

%5Cmathbf%7Bd%7D_1%20%3D%20%5Clangle%201%2C%202%2C%203%20%5Crangle

for

L_1

%5Cmathbf%7Bd%7D_2%20%3D%20%5Clangle%201%2C%200%2C%201%20%5Crangle

for

L_2

Parametric Points:A point on

L_1

P_1(%5Clambda)%20%3D%20(2%20%2B%20%5Clambda%2C%203%20%2B%202%5Clambda%2C%204%20%2B%203%5Clambda)

A point on

L_2

P_2(%5Cmu)%20%3D%20(4%20%2B%20%5Cmu%2C%204%2C%204%20%2B%20%5Cmu)

Vector Between Points:

%5Cbegin%7Baligned%7D%20%5Cmathbf%7BP%7D_1%20-%20%5Cmathbf%7BP%7D_2%20%26%3D%20%5Clangle%20(2%2B%5Clambda)%20-%20(4%2B%5Cmu)%2C%20(3%2B2%5Clambda)%20-%204%2C%20(4%2B3%5Clambda)%20-%20(4%2B%5Cmu)%20%5Crangle%20%5C%5C%20%26%3D%20%5Clangle%20-2%20%2B%20%5Clambda%20-%20%5Cmu%2C%20-1%20%2B%202%5Clambda%2C%203%5Clambda%20-%20%5Cmu%20%5Crangle%20%5Cend%7Baligned%7D

Perpendicular Conditions:

%5Cmathbf%7BP%7D_1%20-%20%5Cmathbf%7BP%7D_2

perpendicular to

%5Cmathbf%7Bd%7D_1

(-2%20%2B%20%5Clambda%20-%20%5Cmu)%20%5Ccdot%201%20%2B%20(-1%20%2B%202%5Clambda)%20%5Ccdot%202%20%2B%20(3%5Clambda%20-%20%5Cmu)%20%5Ccdot%203%20%3D%200

Expanding this:

-2%20%2B%20%5Clambda%20-%20%5Cmu%20-%202%20%2B%204%5Clambda%20%2B%209%5Clambda%20-%203%5Cmu%20%3D%200

Simplifies to:

14%5Clambda%20-%204%5Cmu%20-%204%20%3D%200

Therefore:

7%5Clambda%20-%202%5Cmu%20%3D%202%20%5Cquad%20%5Ctext%7B(Equation%201)%7D

%5Cmathbf%7BP%7D_1%20-%20%5Cmathbf%7BP%7D_2

perpendicular to

%5Cmathbf%7Bd%7D_2

(-2%20%2B%20%5Clambda%20-%20%5Cmu)%20%5Ccdot%201%20%2B%20(-1%20%2B%202%5Clambda)%20%5Ccdot%200%20%2B%20(3%5Clambda%20-%20%5Cmu)%20%5Ccdot%201%20%3D%200

Expanding this:

-2%20%2B%20%5Clambda%20-%20%5Cmu%20%2B%203%5Clambda%20-%20%5Cmu%20%3D%200

Simplifies to:

4%5Clambda%20-%202%5Cmu%20%3D%202%20%5Cquad%20%5Ctext%7B(Equation%202)%7D

Solving Equations:From Equation 2:

2%5Clambda%20-%20%5Cmu%20%3D%201%20%5Cquad%20%5CRightarrow%20%5Cquad%20%5Cmu%20%3D%202%5Clambda%20-%201

Substitute into Equation 1:

7%5Clambda%20-%202(2%5Clambda%20-%201)%20%3D%202

Simplifies to:

7%5Clambda%20-%204%5Clambda%20%2B%202%20%3D%202

3%5Clambda%20%3D%200%20%5Cquad%20%5CRightarrow%20%5Cquad%20%5Clambda%20%3D%200

Substituting

%5Clambda%20%3D%200

into

%5Cmu%20%3D%202%5Clambda%20-%201

%5Cmu%20%3D%20-1

Closest Point on

L_2

:Substitute

%5Cmu%20%3D%20-1

into the equation for

L_2

P_2(-1)%20%3D%20(4%20%2B%20(-1)%2C%204%2C%204%20%2B%20(-1))%20%3D%20(3%2C%204%2C%203)

Thus, the point on

L_2

that is closest to

L_1

(3%2C%204%2C%203)

Three Dimensional Geometry

High-Yield Trend

Questions
2 MCQs

Official Solution

Official Solution

Three Dimensional Geometry

High-Yield Trend

Questions 2 MCQs

Official Solution

Official Solution

Questions
2 MCQs