CBSE Class 12 Maths Exemplar Chapter 11: Three Dimensional Geometry NCERT Solutions

Solution:

Let the position vector of point A be \(\overrightarrow{OA}\). We are given that the magnitude \(|\overrightarrow{OA}| = 10\) units. Let the angles made by \(\overrightarrow{OA}\) with the positive x, y, and z axes be \(\alpha\), \(\beta\), and \(\gamma\) respectively. We are given \(\alpha = 60°\) and \(\beta = 45°\).

We know that the direction cosines of a line satisfy the relation \(l^2 + m^2 + n^2 = 1\), where \(l = \cos \alpha\), \(m = \cos \beta\), and \(n = \cos \gamma\).

Substituting the given values:

$$\backslash cos^2\; 60°\; +\; \backslash cos^2\; 45°\; +\; \backslash cos^2\; \backslash gamma\; =\; 1$$

Calculate the squares of the known cosines:

$$\backslash left(\backslash frac\{1\}\{2\}\backslash right)^2\; +\; \backslash left(\backslash frac\{1\}\{\backslash sqrt\{2\}\}\backslash right)^2\; +\; \backslash cos^2\; \backslash gamma\; =\; 1$$

$$\backslash frac\{1\}\{4\}\; +\; \backslash frac\{1\}\{2\}\; +\; \backslash cos^2\; \backslash gamma\; =\; 1$$

Now, solve for \(\cos^2 \gamma\):

$$\backslash cos^2\; \backslash gamma\; =\; 1\; -\; \backslash left(\backslash frac\{1\}\{4\}\; +\; \backslash frac\{1\}\{2\}\backslash right)$$

$$\backslash cos^2\; \backslash gamma\; =\; 1\; -\; \backslash left(\backslash frac\{1\}\{4\}\; +\; \backslash frac\{2\}\{4\}\backslash right)$$

$$\backslash cos^2\; \backslash gamma\; =\; 1\; -\; \backslash frac\{3\}\{4\}$$

$$\backslash cos^2\; \backslash gamma\; =\; \backslash frac\{1\}\{4\}$$

Taking the square root, we get \(\cos \gamma = \pm \frac{1}{2}\). This implies \(\gamma = 60°\) or \(\gamma = 120°\).

The position vector \(\overrightarrow{OA}\) can be expressed using its magnitude and direction cosines:

$$\backslash overrightarrow\{OA\}\; =\; |\backslash overrightarrow\{OA\}|\; (l\; \backslash hat\{i\}\; +\; m\; \backslash hat\{j\}\; +\; n\; \backslash hat\{k\})$$

Using \(l = \cos 60° = 1/2\), \(m = \cos 45° = 1/\sqrt{2}\), and \(n = \cos 60° = 1/2\) (taking the positive value for \(\cos \gamma\) for simplicity, though \(\pm 1/2\) is possible):

$$\backslash overrightarrow\{OA\}\; =\; 10\; \backslash left(\; \backslash frac\{1\}\{2\}\; \backslash hat\{i\}\; +\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}\; \backslash hat\{j\}\; +\; \backslash frac\{1\}\{2\}\; \backslash hat\{k\}\; \backslash right)$$

Distributing the magnitude:

$$\backslash overrightarrow\{OA\}\; =\; 10\; \backslash times\; \backslash frac\{1\}\{2\}\; \backslash hat\{i\}\; +\; 10\; \backslash times\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}\; \backslash hat\{j\}\; +\; 10\; \backslash times\; \backslash frac\{1\}\{2\}\; \backslash hat\{k\}$$

$$\backslash overrightarrow\{OA\}\; =\; 5\; \backslash hat\{i\}\; +\; \backslash frac\{10\}\{\backslash sqrt\{2\}\}\; \backslash hat\{j\}\; +\; 5\; \backslash hat\{k\}$$

Rationalizing the denominator for the \(j\) component:

$$\backslash overrightarrow\{OA\}\; =\; 5\; \backslash hat\{i\}\; +\; \backslash frac\{10\backslash sqrt\{2\}\}\{2\}\; \backslash hat\{j\}\; +\; 5\; \backslash hat\{k\}$$

$$\backslash overrightarrow\{OA\}\; =\; 5\; \backslash hat\{i\}\; +\; 5\backslash sqrt\{2\}\; \backslash hat\{j\}\; +\; 5\; \backslash hat\{k\}$$

Thus, the position vector of point A is \(5\hat{i} + 5\sqrt{2}\hat{j} + 5\hat{k}\).

Solution:

The vector equation of a line passing through a point with position vector \(\vec{a}\) and parallel to a vector \(\vec{b}\) is given by \(\vec{r} = \vec{a} + \lambda \vec{b}\), where \(\lambda\) is a scalar parameter.

In this problem, the line passes through the point (1, -2, 3). The position vector of this point is \(\vec{a} = 1\hat{i} - 2\hat{j} + 3\hat{k}\).

The line is parallel to the vector \(3\hat{i} - 2\hat{j} + 6\hat{k}\). So, we take \(\vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k}\).

Substituting these into the formula, we get the vector equation of the line:

$$\backslash vec\{r\}\; =\; (\backslash hat\{i\}\; -\; 2\backslash hat\{j\}\; +\; 3\backslash hat\{k\})\; +\; \backslash lambda\; (3\backslash hat\{i\}\; -\; 2\backslash hat\{j\}\; +\; 6\backslash hat\{k\})$$

This equation represents all points on the line. If we let \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\), we can also write the equation in parametric form:

$$x\backslash hat\{i\}\; +\; y\backslash hat\{j\}\; +\; z\backslash hat\{k\}\; =\; (1\; +\; 3\backslash lambda)\backslash hat\{i\}\; +\; (-2\; -\; 2\backslash lambda)\backslash hat\{j\}\; +\; (3\; +\; 6\backslash lambda)\backslash hat\{k\}$$

Equating the components:

\(x = 1 + 3\lambda\)

\(y = -2 - 2\lambda\)

\(z = 3 + 6\lambda\)

The vector equation is \(\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda (3\hat{i} - 2\hat{j} + 6\hat{k})\).

Solution:

First, we need to show that the two lines intersect. Two lines in 3D intersect if and only if the shortest distance between them is zero.

The first line, \(L_1\), has the equation \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}\). This line passes through the point \(P_1(1, 2, 3)\) and has direction ratios \(a_1 = 2, b_1 = 3, c_1 = 4\).

The second line, \(L_2\), has the equation \(\frac{x-4}{5} = \frac{y-1}{2} = \frac{z-0}{1}\) (writing \(z\) as \(\frac{z-0}{1}\)). This line passes through the point \(P_2(4, 1, 0)\) and has direction ratios \(a_2 = 5, b_2 = 2, c_2 = 1\).

The shortest distance between two lines \(\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}\) and \(\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}\) is given by the formula:

Let's calculate the components:

\(x_2 - x_1 = 4 - 1 = 3\)

\(y_2 - y_1 = 1 - 2 = -1\)

\(z_2 - z_1 = 0 - 3 = -3\)

Now, substitute these values and the direction ratios into the determinant for the numerator:

Calculate the determinant:

$$=\; 3((3)(1)\; -\; (4)(2))\; -\; (-1)((2)(1)\; -\; (4)(5))\; +\; (-3)((2)(2)\; -\; (3)(5))$$

$$=\; 3(3\; -\; 8)\; +\; 1(2\; -\; 20)\; -\; 3(4\; -\; 15)$$

$$=\; 3(-5)\; +\; 1(-18)\; -\; 3(-11)$$

$$=\; -15\; -\; 18\; +\; 33$$

$$=\; -33\; +\; 33\; =\; 0$$

Since the numerator of the shortest distance formula is 0, the shortest distance \(d = 0\). This proves that the two lines intersect.

Finding the point of intersection:

Let a point on the first line be \(P\) and a point on the second line be \(Q\). For intersection, \(P = Q\).

Parametric equation for \(L_1\):

\(x = 1 + 2\lambda_1\), \(y = 2 + 3\lambda_1\), \(z = 3 + 4\lambda_1\)

Parametric equation for \(L_2\):

\(x = 4 + 5\lambda_2\), \(y = 1 + 2\lambda_2\), \(z = 0 + 1\lambda_2\)

Equating the coordinates:

1) \(1 + 2\lambda_1 = 4 + 5\lambda_2 \implies 2\lambda_1 - 5\lambda_2 = 3\)

2) \(2 + 3\lambda_1 = 1 + 2\lambda_2 \implies 3\lambda_1 - 2\lambda_2 = -1\)

3) \(3 + 4\lambda_1 = \lambda_2 \implies 4\lambda_1 - \lambda_2 = -3 \implies \lambda_2 = 4\lambda_1 + 3\)

Substitute \(\lambda_2\) from (3) into (1):

\(2\lambda_1 - 5(4\lambda_1 + 3) = 3\)

\(2\lambda_1 - 20\lambda_1 - 15 = 3\)

\(-18\lambda_1 = 18\)

\(\lambda_1 = -1\)

Now find \(\lambda_2\) using \(\lambda_2 = 4\lambda_1 + 3\):

\(\lambda_2 = 4(-1) + 3 = -4 + 3 = -1\)

Let's check if these values satisfy equation (2):

\(3\lambda_1 - 2\lambda_2 = 3(-1) - 2(-1) = -3 + 2 = -1\). This matches equation (2), so our values for \(\lambda_1\) and \(\lambda_2\) are correct.

Now, substitute \(\lambda_1 = -1\) into the parametric equations for \(L_1\) to find the point of intersection:

\(x = 1 + 2(-1) = 1 - 2 = -1\)

\(y = 2 + 3(-1) = 2 - 3 = -1\)

\(z = 3 + 4(-1) = 3 - 4 = -1\)

Alternatively, substitute \(\lambda_2 = -1\) into the parametric equations for \(L_2\):

\(x = 4 + 5(-1) = 4 - 5 = -1\)

\(y = 1 + 2(-1) = 1 - 2 = -1\)

\(z = 0 + 1(-1) = -1\)

Both lines give the same point of intersection. Therefore, the point of intersection is (-1, -1, -1).