Class 12 Mathematics Chapter 11 Three Dimensional Geometry

Key Concept: Cartesian Conversion

a) $$\frac{x – 4}{2} = \frac{y – 7}{-3} = \frac{z – 2}{1}$$
[Solution Description] Given the vector equation $$\mathbf{r} = 4\mathbf{i} + 7\mathbf{j} + 2\mathbf{k} + \lambda(2\mathbf{i} – 3\mathbf{j} + \mathbf{k})$$, we extract the parametric equations: $$x = 4 + 2\lambda$$, $$y = 7 – 3\lambda$$, $$z = 2 + \lambda$$. To convert to Cartesian form, solve each parametric equation for $$\lambda$$:
\begin{align*}
\frac{x – 4}{2} &= \frac{y – 7}{-3} = \frac{z – 2}{1}.
\end{align*}

Key Concept: Direction Cosines and Ratios, Multi-step Solutions

a) $$\left(\frac{1}{3}, \frac{-2}{3}, \frac{2}{3}\right)$$
[Solution Description] To convert direction ratios $$(a, b, c)$$ to direction cosines $$(l, m, n)$$, use the formulas:
$$
l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}
$$
With $$(a, b, c) = (1, -2, 2)$$, compute the magnitude:
$$
\sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
$$

Therefore:
$$
l = \frac{1}{3}, \quad m = \frac{-2}{3}, \quad n = \frac{2}{3}
$$

So, the direction cosines are $$\left(\frac{1}{3}, \frac{-2}{3}, \frac{2}{3}\right)$$.

Key Concept: Vector Equation

c) $$\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}$$
[Solution Description]
The vector equation of a line through a point with position vector $$\mathbf{a}$$ and parallel to vector $$\mathbf{b}$$ is given by:
$$
\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}
$$
where $$\lambda$$ is a scalar parameter.

Key Concept: Unit Vector along Shortest Distance

d) $$\frac{\mathbf{b_1} \times \mathbf{b_2}}{|\mathbf{b_1} \times \mathbf{b_2}|}$$
[Solution Description] The unit vector along the shortest distance between skew lines with given direction vectors is expressed as $$\hat{n} = \frac{\mathbf{b_1} \times \mathbf{b_2}}{|\mathbf{b_1} \times \mathbf{b_2}|}$$.

Key Concept: Basic Line Equation

d) $$\frac{x}{4} = \frac{y}{-3} = \frac{z}{5}$$
[Solution Description] The equation of a line passing through the point $$(x_0, y_0, z_0)$$ and parallel to the vector $$\vec{a} = a\hat{i} + b\hat{j} + c\hat{k}$$ is given by:

$$
\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}
$$

Substituting the given values, we have:

$$
\frac{x}{4} = \frac{y}{-3} = \frac{z}{5}
$$

So, the equation of the line is $$\frac{x}{4} = \frac{y}{-3} = \frac{z}{5}$$.

Angle Calculation

a) $$\theta = \cos^{-1}\left(\frac{8}{\sqrt{294}}\right)$$

[Solution Description] The angle $$\theta$$ between two vectors can be found using:

$$\cos \theta = \frac{b_1 \cdot b_2}{|b_1||b_2|}$$

Compute the dot product:

$$b_1 \cdot b_2 = (2)(-1) + (3)(4) + (-1)(2) = -2 + 12 – 2 = 8$$

Calculate magnitudes:

$$|b_1| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$$

$$|b_2| = \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$$

Therefore,

$$\cos \theta = \frac{8}{\sqrt{14} \times \sqrt{21}} = \frac{8}{\sqrt{294}}$$

The angle $$\theta$$ is:

$$\theta = \cos^{-1}\left(\frac{8}{\sqrt{294}}\right)$$

We provide it without simplification.

Key Concept: Basic Formula

a) $$ \cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} $$
[Solution Description] The cosine of the angle between two lines with direction ratios can be calculated using the formula:

$$
\cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}
$$

Here, $$a_1, b_1, c_1$$ are the direction ratios of the first line and $$a_2, b_2, c_2$$ are the direction ratios of the second line.

Key Concept: Complex Distance Calculation, Real World Application

a) $$\frac{6}{\sqrt{69}}$$
[Solution Description] Utilize the shortest distance formula for skew lines:
$$ d = \frac{|(\mathbf{a}_2 – \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}, $$
where $$\mathbf{a}_1 = 3\hat{i} + 2\hat{j}$$, $$\mathbf{a}_2 = 5\hat{i} + \hat{j}$$, $$\mathbf{b}_1 = 4\hat{i} – \hat{j} + 2\hat{k}$$, and $$\mathbf{b}_2 = 3\hat{i} + \hat{j} + 2\hat{k}$$.

Calculate the cross product of direction vectors:
\begin{align*}
\mathbf{b}_1 \times \mathbf{b}_2 &=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
4 & -1 & 2 \\
3 & 1 & 2
\end{vmatrix} = \hat{i}((-1)(2) – (2)(1)) – \hat{j}(4(2) – (3)(2)) + \hat{k}(4(1) – 3(-1)) \\
&= \hat{i}(-2 – 2) – \hat{j}(8 – 6) + \hat{k}(4 + 3) = -4\hat{i} – 2\hat{j} + 7\hat{k}.
\end{align*}

Magnitude of the cross product:
\begin{align*}
|\mathbf{b}_1 \times \mathbf{b}_2| = \sqrt{(-4)^2 + (-2)^2 + 7^2} = \sqrt{16 + 4 + 49} = \sqrt{69}.
\end{align*}

Calculate $$(\mathbf{a}_2 – \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)$$:
\begin{align*}
(\mathbf{a}_2 – \mathbf{a}_1) &= (5\hat{i} + \hat{j}) – (3\hat{i} + 2\hat{j}) = 2\hat{i} – \hat{j}, \\
(\mathbf{a}_2 – \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2) &= (2\hat{i} – \hat{j}) \cdot (-4\hat{i} – 2\hat{j} + 7\hat{k}) \\
&= 2(-4) + (-1)(-2) + 0 = -8 + 2 = -6.
\end{align*}

Finally, compute the shortest distance:
\begin{align*}
d = \frac{| -6 |}{\sqrt{69}} = \frac{6}{\sqrt{69}}.
\end{align*}

Thus, the shortest distance is $$\frac{6}{\sqrt{69}}$$.

Key concept: Conversion to Cartesian

b) $$\frac{x – 7}{6} = \frac{y + 1}{3} = \frac{z – 2}{-4}$$

[Solution Description]

Given the vector equation:

$$\mathbf{r} = (7, -1, 2) + \lambda(6, 3, -4)$$

This corresponds to the parametric equations:

$$x = 7 + 6\lambda$$

$$y = -1 + 3\lambda$$

$$z = 2 – 4\lambda$$

To convert it to the Cartesian form, eliminate $$\lambda$$ from these equations.

From the first equation:

$$\lambda = \frac{x – 7}{6}$$

Substitute in the second equation:

$$\frac{y + 1}{3} = \frac{x – 7}{6}$$

And from the third equation:

$$\frac{z – 2}{-4} = \frac{x – 7}{6}$$

Therefore, the Cartesian equation is:

$$\frac{x – 7}{6} = \frac{y + 1}{3} = \frac{z – 2}{-4}$$

This represents the Cartesian form of the given vector equation.

Key Concept: Basic Concept

a) $$\left(\frac{1}{2}, \frac{\sqrt{2}}{2}, 0\right)$$
[Solution Description] The direction cosines of a line are given by $$l = \cos \alpha$$, $$m = \cos \beta$$, $$n = \cos \gamma$$. For $$\alpha = 60^\circ$$, $$\beta = 45^\circ$$, and $$\gamma = 90^\circ$$:

$$
l = \cos 60^\circ = \frac{1}{2}, \quad m = \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad n = \cos 90^\circ = 0.
$$

Thus, the direction cosines are $$\left(\frac{1}{2}, \frac{\sqrt{2}}{2}, 0\right)$$.

Key concept: Triangle Sides

b) $$\left(\frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}, \frac{2}{\sqrt{29}}\right)$$

[Solution Description]

The direction ratios for the line joining points A and B are obtained by finding the differences in their coordinates: $$(x_2 – x_1, y_2 – y_1, z_2 – z_1) = (4-1, 2-(-2), 5-3) = (3, 4, 2)$$. The magnitude is $$PQ = \sqrt{3^2 + 4^2 + 2^2} = \sqrt{29}$$. Thus, the direction cosines are:

$$l = \frac{3}{\sqrt{29}}, \quad m = \frac{4}{\sqrt{29}}, \quad n = \frac{2}{\sqrt{29}}.$$

Hence, the direction cosines are $$\left(\frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}, \frac{2}{\sqrt{29}}\right)$$.

Key Concept: Basic Formula

a) $$\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{3}$$
[Solution Description] The Cartesian form of the line is given by $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$$. Substituting $$(x_1, y_1, z_1) = (3, 4, 5)$$ and direction ratios $$(a, b, c) = (1, 2, 3)$$, we get:
$$
\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{3}.
$$

Key concept: Angle Calculation

c) $$180^\circ$$

[Solution Description]

To find the angle $$\theta$$ between two lines given their direction ratios, we use the formula:

$$\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}$$

Substituting the given direction ratios: $$(a_1, b_1, c_1) = (3, -4, 5)$$ and $$(a_2, b_2, c_2) = (-6, 8, -10)$$:

$$\cos \theta = \frac{3(-6) + (-4)(8) + 5(-10)}{\sqrt{3^2 + (-4)^2 + 5^2} \cdot \sqrt{(-6)^2 + 8^2 + (-10)^2}}$$

Calculating the numerator:

$$3(-6) + (-4)(8) + 5(-10) = -18 – 32 – 50 = -100$$

Calculating the denominator:

$$\sqrt{9 + 16 + 25} \cdot \sqrt{36 + 64 + 100} = \sqrt{50} \cdot \sqrt{200} = \sqrt{10000} = 100$$

Thus,

$$\cos \theta = \frac{-100}{100} = -1$$

Therefore, the angle $$\theta$$ is $$180^\circ$$.

Key concept: Shortest Distance

c) $$\frac{40}{\sqrt{195}}$$

[Solution Description] To find the shortest distance between two skew lines, we use the formula:

$$d = \frac{|(\mathbf{a}_2 – \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}$$

Given:

$$\mathbf{a}_1 = \langle 2, 3, 1 \rangle$$,

$$\mathbf{a}_2 = \langle 4, -1, 3 \rangle$$,

$$\mathbf{b}_1 = \langle 3, -1, 2 \rangle$$,

$$\mathbf{b}_2 = \langle -1, 2, 3 \rangle$$.

Calculate $$\mathbf{a}_2 – \mathbf{a}_1 = \langle 4 – 2, -1 – 3, 3 – 1 \rangle = \langle 2, -4, 2 \rangle$$.

Next, calculate $$\mathbf{b}_1 \times \mathbf{b}_2$$ using cross product:

$$\mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 -1  -2 \\ -1  -2  -3 \end{vmatrix}$$

Calculate determinants:

$$= \mathbf{i}( (-1)(3) – (2)(2) ) – \mathbf{j}( (3)(3) – (2)(-1) ) + \mathbf{k}( (3)(2) – (-1)(-1) )$$

$$= \mathbf{i}(-3 – 4) – \mathbf{j}(9 + 2) + \mathbf{k}(6 – 1)$$

$$= -7\mathbf{i} – 11\mathbf{j} + 5\mathbf{k}$$

Magnitude of $$\mathbf{b}_1 \times \mathbf{b}_2$$:

$$|\mathbf{b}_1 \times \mathbf{b}_2| = \sqrt{(-7)^2 + (-11)^2 + 5^2} = \sqrt{49 + 121 + 25} = \sqrt{195}$$

Dot Product $$(\mathbf{a}_2 – \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)$$:

$$\langle 2, -4, 2 \rangle \cdot \langle -7, -11, 5 \rangle = (2)(-7) + (-4)(-11) + (2)(5) = -14 + 44 + 10 = 40$$

Shortest distance d becomes:

$$d = \frac{|40|}{\sqrt{195}} = \frac{40}{\sqrt{195}}$$

Key Concept: Unit Vector Calculation

a) $$\frac{\mathbf{b}_1 \times \mathbf{b}_2}{|\mathbf{b}_1 \times \mathbf{b}_2|}$$
[Solution Description] To find the unit vector perpendicular to both $$\mathbf{b}_1$$ and $$\mathbf{b}_2$$, calculate their cross product $$\mathbf{b}_1 \times \mathbf{b}_2$$, then find the unit vector by dividing by the magnitude of this cross product.
$$
\text{Unit vector} = \frac{\mathbf{b}_1 \times \mathbf{b}_2}{|\mathbf{b}_1 \times \mathbf{b}_2|}
$$

Key Concept: Formula Application

b) $$\frac{4}{13}, \frac{-3}{13}, \frac{12}{13}$$
[Solution Description]
Calculate the magnitude of the direction ratios:
$$
\sqrt{4^2 + (-3)^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13
$$

The direction cosines are:
\begin{align*}
l &= \frac{4}{13}, \\
m &= \frac{-3}{13}, \\
n &= \frac{12}{13}
\end{align*}
So the direction cosines are $$\left(\frac{4}{13}, \frac{-3}{13}, \frac{12}{13}\right)$$.

Key Concept: Axis Cosines

b) $$l = 0, m = 1, n = 0$$
[Solution Description] The direction cosines of the y-axis correspond to it being aligned along the y-direction, which means it makes an angle of 90° with the x-axis, 0° with the y-axis, and 90° with the z-axis. Hence, the direction cosines are:

$$
l = \cos 90^\circ = 0, \quad m = \cos 0^\circ = 1, \quad n = \cos 90^\circ = 0
$$

Therefore, the direction cosines of the y-axis are $$(0, 1, 0)$$.