Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.6

Question 1. Find a parametric form of vector equation of a plane which is at a distance of 7 units from the origin having 3, -4, 5 as direction ratios of a normal to it.
Answer:
Given that the distance of the plane from the origin, \( p = 7 \).
The direction ratios of the normal to the plane are 3, -4, 5.
So, the normal vector is \( \vec{d} = 3\hat{i} - 4\hat{j} + 5\hat{k} \).
First, find the magnitude of the normal vector:
\( |\vec{d}| = \sqrt{3^2 + (-4)^2 + 5^2} \)
\( = \sqrt{9 + 16 + 25} \)
\( = \sqrt{50} \)
\( = \sqrt{25 \times 2} \)
\( = 5\sqrt{2} \)
Now, find the unit normal vector \( \hat{n} \):
\( \hat{n} = \frac{\vec{d}}{|\vec{d}|} \)
\( \hat{n} = \frac{3\hat{i} - 4\hat{j} + 5\hat{k}}{5\sqrt{2}} \)
The parametric form of the vector equation of a plane is given by \( \vec{r} \cdot \hat{n} = p \).
Substitute the values of \( \hat{n} \) and \( p \):
\( \vec{r} \cdot \left( \frac{3\hat{i} - 4\hat{j} + 5\hat{k}}{5\sqrt{2}} \right) = 7 \)
This is the required parametric form of the vector equation of the plane. This equation shows the relationship between any point on the plane and its normal vector.
In simple words: We start with the distance from the origin and the normal vector's direction. Then, we find the unit normal vector by dividing the normal vector by its length. Finally, we put these into a standard formula to get the plane's equation.

🎯 Exam Tip: Remember to always normalize the normal vector to get the unit normal vector \( \hat{n} \) before using the formula \( \vec{r} \cdot \hat{n} = p \).

Question 2. Find the direction cosines of the normal to the plane \( 12x + 3y – 4z = 65 \). Also find the non-parametric form of vector equation of a plane and the length of the perpendicular to the plane from the origin.
Answer:
The given equation of the plane is \( 12x + 3y - 4z = 65 \).
Comparing this with the standard form \( Ax + By + Cz = D \), we have:
\( A = 12 \), \( B = 3 \), \( C = -4 \), \( D = 65 \).
The normal vector to the plane is \( \vec{d} = 12\hat{i} + 3\hat{j} - 4\hat{k} \).
The length of the normal vector is:
\( |\vec{d}| = \sqrt{12^2 + 3^2 + (-4)^2} \)
\( = \sqrt{144 + 9 + 16} \)
\( = \sqrt{169} \)
\( = 13 \)
The direction cosines of the normal are \( \left( \frac{A}{|\vec{d}|}, \frac{B}{|\vec{d}|}, \frac{C}{|\vec{d}|} \right) \).
Direction cosines: \( \left( \frac{12}{13}, \frac{3}{13}, \frac{-4}{13} \right) \).
For the non-parametric form of the vector equation, we use \( \vec{r} \cdot \vec{d} = D \).
So, \( \vec{r} \cdot (12\hat{i} + 3\hat{j} - 4\hat{k}) = 65 \). This equation uses the normal vector directly.
The length of the perpendicular from the origin to the plane is given by \( \frac{|D|}{|\vec{d}|} \).
Length of perpendicular \( = \frac{|65|}{13} = 5 \) units.
In simple words: From the plane's equation, we find the normal vector and its length. Then, we calculate the direction cosines by dividing each part of the normal vector by its length. The non-parametric vector equation uses the normal vector as is. Finally, we find the distance from the origin by dividing the constant term by the normal vector's length.

🎯 Exam Tip: Always remember that the coefficients of x, y, z in the Cartesian equation of a plane directly give the direction ratios of its normal vector. The length of the perpendicular from the origin is always positive.

Question 3. Find the vector and Cartesian equations of the plane passing through the point with position vector \( 2\hat { i } + 6\hat { j } + 3\hat { k } \) and normal to the vector \( \hat { i } + 3\hat { j } + 5\hat { k } \).
Answer:
Let the position vector of the point through which the plane passes be \( \vec{a} \).
\( \vec{a} = 2\hat{i} + 6\hat{j} + 3\hat{k} \).
Let the normal vector to the plane be \( \vec{n} \).
\( \vec{n} = \hat{i} + 3\hat{j} + 5\hat{k} \).
The vector equation of a plane passing through a point \( \vec{a} \) and normal to a vector \( \vec{n} \) is given by \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \).
Substitute the values of \( \vec{a} \) and \( \vec{n} \):
\( (\vec{r} - (2\hat{i} + 6\hat{j} + 3\hat{k})) \cdot (\hat{i} + 3\hat{j} + 5\hat{k}) = 0 \).
This is the vector equation of the plane. To simplify, we can write it as \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \).
First, calculate \( \vec{a} \cdot \vec{n} \):
\( \vec{a} \cdot \vec{n} = (2\hat{i} + 6\hat{j} + 3\hat{k}) \cdot (\hat{i} + 3\hat{j} + 5\hat{k}) \)
\( = (2)(1) + (6)(3) + (3)(5) \)
\( = 2 + 18 + 15 \)
\( = 35 \).
So, the vector equation is \( \vec{r} \cdot (\hat{i} + 3\hat{j} + 5\hat{k}) = 35 \). This equation helps visualize the plane's orientation in space.

To find the Cartesian equation, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Substitute \( \vec{r} \) into the vector equation:
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 3\hat{j} + 5\hat{k}) = 35 \)
\( (x)(1) + (y)(3) + (z)(5) = 35 \)
\( x + 3y + 5z = 35 \).
This is the Cartesian equation of the plane.
In simple words: We are given a point the plane passes through and a vector normal to the plane. We use a formula involving these two to get the vector equation. Then, we substitute \( \vec{r} \) with \( x\hat{i} + y\hat{j} + z\hat{k} \) to easily find the Cartesian equation.

🎯 Exam Tip: Remember the two forms: \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \) for the vector equation and \( x_1 + y_1 + z_1 = D \) for the Cartesian equation. Be careful with the dot product calculations.

Question 4. A plane passes through the point (-1, 1, 2) and the normal to the plane of magnitude \( 3\sqrt{3} \) makes equal acute angles with the co-ordinate axis. Find the equation of the plane.
Answer:
Let the angles made by the normal with the co-ordinate axes be \( \alpha, \beta, \gamma \).
Given that the normal makes equal acute angles with the co-ordinate axes, so \( \alpha = \beta = \gamma \).
The direction cosines are \( l = \cos \alpha \), \( m = \cos \beta \), \( n = \cos \gamma \).
We know that \( l^2 + m^2 + n^2 = 1 \).
So, \( \cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \).
\( 3\cos^2 \alpha = 1 \)
\( \cos^2 \alpha = \frac{1}{3} \)
\( \cos \alpha = \pm \frac{1}{\sqrt{3}} \).
Since the angles are acute, we take the positive value: \( \cos \alpha = \frac{1}{\sqrt{3}} \).
Thus, the direction cosines are \( \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \). These cosines define the orientation of the normal vector.
The magnitude of the normal vector \( |\vec{n}| = 3\sqrt{3} \).
The normal vector \( \vec{n} = |\vec{n}| (l\hat{i} + m\hat{j} + n\hat{k}) \).
\( \vec{n} = 3\sqrt{3} \left( \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \right) \)
\( \vec{n} = 3\hat{i} + 3\hat{j} + 3\hat{k} \).
The plane passes through the point \( (-1, 1, 2) \). So, \( \vec{a} = -\hat{i} + \hat{j} + 2\hat{k} \).
The equation of the plane is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \).
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
\( ((x\hat{i} + y\hat{j} + z\hat{k}) - (-\hat{i} + \hat{j} + 2\hat{k})) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = 0 \)
\( ((x+1)\hat{i} + (y-1)\hat{j} + (z-2)\hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = 0 \).
Take the dot product:
\( 3(x+1) + 3(y-1) + 3(z-2) = 0 \).
Divide by 3:
\( (x+1) + (y-1) + (z-2) = 0 \).
\( x + 1 + y - 1 + z - 2 = 0 \).
\( x + y + z - 2 = 0 \).
\( x + y + z = 2 \).
This is the Cartesian equation of the plane. This equation shows all points (x, y, z) that lie on this specific plane.
The vector equation can also be written as \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2 \).
In simple words: First, we find the normal vector by knowing it makes equal angles with the axes and its magnitude. Then, we use the point the plane passes through and the normal vector in a formula to get the plane's equation.

🎯 Exam Tip: When a normal makes equal acute angles with the axes, its direction cosines are \( \pm \frac{1}{\sqrt{3}} \). For acute angles, always choose the positive values.

Question 5. Find the intercepts cut off by the plane \( \vec{r} \cdot (6\hat{i} + 4\hat{j} - 3\hat{k}) = 12 \) on the co-ordinate axes.
Answer:
The given vector equation of the plane is \( \vec{r} \cdot (6\hat{i} + 4\hat{j} - 3\hat{k}) = 12 \).
To find the intercepts, we first convert this into its Cartesian form.
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
So, \( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (6\hat{i} + 4\hat{j} - 3\hat{k}) = 12 \).
Taking the dot product:
\( 6x + 4y - 3z = 12 \).
This is the Cartesian equation of the plane. Now, to find the intercepts, we divide the entire equation by the constant on the right side (12) to get it in the intercept form \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
Divide by 12:
\( \frac{6x}{12} + \frac{4y}{12} - \frac{3z}{12} = \frac{12}{12} \)
\( \frac{x}{2} + \frac{y}{3} + \frac{z}{-4} = 1 \).
From this form, we can directly read the intercepts:
The x-intercept is \( a = 2 \).
The y-intercept is \( b = 3 \).
The z-intercept is \( c = -4 \).
These intercepts are the points where the plane crosses the x, y, and z axes, respectively. An understanding of intercepts is crucial in graphing planes.
In simple words: We take the plane's vector equation and turn it into a regular equation with x, y, and z. Then, we divide the whole equation by the constant on the right side. The numbers under x, y, and z in the new equation are our intercepts.

🎯 Exam Tip: The intercept form of a plane equation is \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \), where a, b, c are the x, y, z intercepts respectively. Ensure the right-hand side is always 1 for this formula to work correctly.

Question 6. If a plane meets the co-ordinate axes at A, B, C such that the centroid of the triangle ABC is the point (u, v, w), find the equation of the plane.
Answer:
Let the intercepts cut off by the plane on the co-ordinate axes be \( a \), \( b \), and \( c \).
Then the points where the plane meets the axes are:
A = \( (a, 0, 0) \)
B = \( (0, b, 0) \)
C = \( (0, 0, c) \)
The equation of the plane in intercept form is \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \). This form directly shows where the plane intersects the axes.
The centroid of the triangle ABC is given by the formula:
\( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \).
Substitute the coordinates of A, B, C:
Centroid \( = \left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) \)
Centroid \( = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right) \).
We are given that the centroid is \( (u, v, w) \).
So, \( u = \frac{a}{3} \implies a = 3u \).
\( v = \frac{b}{3} \implies b = 3v \).
\( w = \frac{c}{3} \implies c = 3w \).
Now, substitute these values of \( a, b, c \) back into the intercept form of the plane equation:
\( \frac{x}{3u} + \frac{y}{3v} + \frac{z}{3w} = 1 \).
Multiply the entire equation by 3 to simplify:
\( \frac{3x}{3u} + \frac{3y}{3v} + \frac{3z}{3w} = 3 \times 1 \)
\( \frac{x}{u} + \frac{y}{v} + \frac{z}{w} = 3 \).
This is the required equation of the plane. Understanding how geometric properties like centroids relate to algebraic equations is fundamental in coordinate geometry.
In simple words: We assume the plane cuts the axes at points (a,0,0), (0,b,0), and (0,0,c). Then we use the formula for a triangle's center point (centroid) and what the problem tells us about it. This helps us find 'a', 'b', and 'c' in terms of 'u', 'v', 'w', which we then put into the standard plane equation.

🎯 Exam Tip: Remember that the centroid of a triangle with vertices \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), \( (x_3, y_3, z_3) \) is \( \left( \frac{\Sigma x}{3}, \frac{\Sigma y}{3}, \frac{\Sigma z}{3} \right) \). Use this relation to connect intercepts to the centroid's coordinates.