GSEB Class 12 Maths Solutions Chapter 11 ત્રિપરિમાણીય ભૂમિતિ Exercise 11.3

 

Question 1. નીચેના પૈકી દરેક પ્રશ્નમાં સમતલના અભિલંબની દિક્કોસાઇન અને સમતલનું ઊગમબિંદુથી અંતર મેળવો.
(a) \( z = 2 \)
(b) \( x + y + z = 1 \)
(c) \( 2x + 3y - z = 5 \)
(d) \( 5y + 8 = 0 \)
Answer:
(a) For the plane \( z = 2 \):
The equation can be written as \( 0x + 0y + 1z - 2 = 0 \).
Comparing this with the general plane equation \( ax + by + cz + d = 0 \), we get \( a = 0, b = 0, c = 1, d = -2 \).
The direction cosines (d.c.'s) of the normal to the plane are \( l, m, n \):
\( l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{0}{\sqrt{0^2+0^2+1^2}} = \frac{0}{1} = 0 \)
\( m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{0}{\sqrt{0^2+0^2+1^2}} = \frac{0}{1} = 0 \)
\( n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{0^2+0^2+1^2}} = \frac{1}{1} = 1 \)
Thus, the direction cosines of the normal are 0, 0, 1.
The distance of the plane from the origin O(0, 0, 0) is given by:
\( D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \)
\( = \frac{|0(0)+0(0)+1(0)+(-2)|}{\sqrt{0^2+0^2+1^2}} \)
\( = \frac{|-2|}{\sqrt{1}} = \frac{2}{1} = 2 \)
The distance is 2 units.
(b) For the plane \( x + y + z = 1 \):
The equation can be written as \( 1x + 1y + 1z - 1 = 0 \).
Comparing this with \( ax + by + cz + d = 0 \), we get \( a = 1, b = 1, c = 1, d = -1 \).
The direction cosines of the normal to the plane are \( l, m, n \):
\( l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{1^2+1^2+1^2}} = \frac{1}{\sqrt{3}} \)
\( m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{1^2+1^2+1^2}} = \frac{1}{\sqrt{3}} \)
\( n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{1^2+1^2+1^2}} = \frac{1}{\sqrt{3}} \)
Thus, the direction cosines of the normal are \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \).
The distance of the plane from the origin O(0, 0, 0) is given by:
\( D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \)
\( = \frac{|1(0)+1(0)+1(0)+(-1)|}{\sqrt{1^2+1^2+1^2}} \)
\( = \frac{|-1|}{\sqrt{3}} = \frac{1}{\sqrt{3}} \)
The distance is \( \frac{1}{\sqrt{3}} \) units.
(c) For the plane \( 2x + 3y - z = 5 \):
The equation can be written as \( 2x + 3y - 1z - 5 = 0 \).
Comparing this with \( ax + by + cz + d = 0 \), we get \( a = 2, b = 3, c = -1, d = -5 \).
The direction cosines of the normal to the plane are \( l, m, n \):
\( l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{2}{\sqrt{2^2+3^2+(-1)^2}} = \frac{2}{\sqrt{4+9+1}} = \frac{2}{\sqrt{14}} \)
\( m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{3}{\sqrt{2^2+3^2+(-1)^2}} = \frac{3}{\sqrt{4+9+1}} = \frac{3}{\sqrt{14}} \)
\( n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{-1}{\sqrt{2^2+3^2+(-1)^2}} = \frac{-1}{\sqrt{4+9+1}} = \frac{-1}{\sqrt{14}} \)
Thus, the direction cosines of the normal are \( \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}} \).
The distance of the plane from the origin O(0, 0, 0) is given by:
\( D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \)
\( = \frac{|2(0)+3(0)+(-1)(0)+(-5)|}{\sqrt{2^2+3^2+(-1)^2}} \)
\( = \frac{|-5|}{\sqrt{14}} = \frac{5}{\sqrt{14}} \)
The distance is \( \frac{5}{\sqrt{14}} \) units.
(d) For the plane \( 5y + 8 = 0 \):
The equation can be written as \( 0x + 5y + 0z + 8 = 0 \).
Comparing this with \( ax + by + cz + d = 0 \), we get \( a = 0, b = 5, c = 0, d = 8 \).
The direction cosines of the normal to the plane are \( l, m, n \):
\( l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{0}{\sqrt{0^2+5^2+0^2}} = \frac{0}{5} = 0 \)
\( m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{5}{\sqrt{0^2+5^2+0^2}} = \frac{5}{5} = 1 \)
\( n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{0}{\sqrt{0^2+5^2+0^2}} = \frac{0}{5} = 0 \)
Thus, the direction cosines of the normal are 0, 1, 0.
The distance of the plane from the origin O(0, 0, 0) is given by:
\( D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \)
\( = \frac{|0(0)+5(0)+0(0)+8|}{\sqrt{0^2+5^2+0^2}} \)
\( = \frac{|8|}{\sqrt{25}} = \frac{8}{5} \)
The distance is \( \frac{8}{5} \) units.
In simple words: For each plane, we first write its equation in the standard form \( ax + by + cz + d = 0 \). Then, we find the direction cosines of the normal vector by dividing each coefficient \( (a, b, c) \) by the magnitude of the normal vector \( \sqrt{a^2+b^2+c^2} \). Finally, we calculate the perpendicular distance from the origin using the distance formula, ensuring to take the absolute value of the numerator.

Exam Tip: Remember that the constant 'd' in the plane equation \(ax+by+cz+d=0\) is crucial for calculating the distance from the origin. Always make sure to keep track of the signs of 'a', 'b', 'c', and 'd' when applying the distance formula.

 

Question 2. ઊગમબિંદુથી 7 એકમ અંતરે આવેલા અને જેનો અભિલંબ સદિશ \( 3\hat{i} + 5\hat{j} – 6\hat{k} \) હોય તેવા સમતલનું સંદેશ સમીકરણ શોધો.
Answer:
The given normal vector is \( \vec{n} = 3\hat{i} + 5\hat{j} – 6\hat{k} \).
First, we calculate the magnitude of the normal vector:
\( |\vec{n}| = \sqrt{3^2 + 5^2 + (-6)^2} \)
\( = \sqrt{9 + 25 + 36} = \sqrt{70} \)
Next, we find the unit normal vector \( \hat{n} \):
\( \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{3\hat{i} + 5\hat{j} – 6\hat{k}}{\sqrt{70}} \)
The distance of the plane from the origin is given as \( d = 7 \) units.
The vector equation of a plane at a distance \( d \) from the origin and with unit normal vector \( \hat{n} \) is \( \vec{r} \cdot \hat{n} = d \).
Substituting the values, we get the equation of the plane:
\( \vec{r} \cdot \left(\frac{3\hat{i} + 5\hat{j} – 6\hat{k}}{\sqrt{70}}\right) = 7 \)
In simple words: We are given the normal vector and the distance from the origin. First, we find the unit normal vector by dividing the normal vector by its length. Then, we use the standard formula \( \vec{r} \cdot \hat{n} = d \) to write the vector equation of the plane.

Exam Tip: Remember the standard form of the vector equation of a plane: \( \vec{r} \cdot \hat{n} = d \), where \( \hat{n} \) is the unit normal vector and \( d \) is the perpendicular distance from the origin.

 

Question 3. નીચેના પૈકી પ્રત્યેક સમતલનું કાર્તેઝિય સમીકરણ શોધો :
(a) \( \vec{r} \cdot (\hat{i} + \hat{j} – \hat{k}) = 2 \)
(b) \( \vec{r} \cdot (2\hat{i} + 3\hat{j} -4\hat{k}) = 1 \)
(c) \( \vec{r} \cdot ((s - 2t)\hat{i} + (3 – t)\hat{j} + (2s + t)\hat{k} = 15 \)
Answer:
(a) For the plane \( \vec{r} \cdot (\hat{i} + \hat{j} – \hat{k}) = 2 \):
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Substitute this into the vector equation:
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} – \hat{k}) = 2 \)
Performing the dot product, we get:
\( x(1) + y(1) + z(-1) = 2 \)
\( x + y - z = 2 \)
This is the Cartesian equation of the given plane.
(b) For the plane \( \vec{r} \cdot (2\hat{i} + 3\hat{j} -4\hat{k}) = 1 \):
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Substitute this into the vector equation:
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 3\hat{j} -4\hat{k}) = 1 \)
Performing the dot product, we get:
\( x(2) + y(3) + z(-4) = 1 \)
\( 2x + 3y - 4z = 1 \)
This is the Cartesian equation of the given plane.
(c) For the plane \( \vec{r} \cdot ((s - 2t)\hat{i} + (3 – t)\hat{j} + (2s + t)\hat{k}) = 15 \):
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Substitute this into the vector equation:
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot ((s - 2t)\hat{i} + (3 – t)\hat{j} + (2s + t)\hat{k}) = 15 \)
Performing the dot product, we get:
\( x(s - 2t) + y(3 – t) + z(2s + t) = 15 \)
This is the Cartesian equation of the given plane.
In simple words: To change a plane's vector equation \( \vec{r} \cdot \vec{n} = d \) into its Cartesian form, we simply replace the vector \( \vec{r} \) with \( x\hat{i} + y\hat{j} + z\hat{k} \). Then, we perform the dot product of this position vector with the normal vector \( \vec{n} \) and set it equal to \( d \).

Exam Tip: To convert a vector equation \( \vec{r} \cdot \vec{n} = d \) to Cartesian form, simply replace \( \vec{r} \) with \( x\hat{i} + y\hat{j} + z\hat{k} \) and perform the dot product.

 

Question 4. નીચેના પૈકી પ્રત્યેક પ્રશ્નમાં ઊગમબિંદુથી સમતલ પર દોરેલા લંબના લંબપાદના યામ શોધો :
(a) \( 2x + 3y + 4z = 12 \)
(b) \( 3y + 4z = 6 \)
(c) \( x + y + z = 1 \)
(d) \( 5y + 8 = 0 \)
Answer:
(a) For the plane \( 2x + 3y + 4z = 12 \):
The given plane equation is \( 2x + 3y + 4z - 12 = 0 \).
The direction ratios of the normal to this plane are 2, 3, 4.
The equation of the line passing through the origin O(0, 0, 0) and parallel to this normal (which is the perpendicular from the origin to the plane) is:
\( \frac{x-0}{2}=\frac{y-0}{3}=\frac{z-0}{4} = \lambda \)
From this, we get \( x = 2\lambda, y = 3\lambda, z = 4\lambda \).
Let M be the foot of the perpendicular from the origin O to the plane. The coordinates of M are \( (2\lambda, 3\lambda, 4\lambda) \).
Since point M lies on the plane \( 2x + 3y + 4z - 12 = 0 \), its coordinates must satisfy the plane's equation:
\( 2(2\lambda) + 3(3\lambda) + 4(4\lambda) - 12 = 0 \)
\( 4\lambda + 9\lambda + 16\lambda - 12 = 0 \)
\( 29\lambda - 12 = 0 \)
\( \implies \lambda = \frac{12}{29} \)
Now, substitute the value of \( \lambda \) back into the coordinates of M:
\( M = \left(2\left(\frac{12}{29}\right), 3\left(\frac{12}{29}\right), 4\left(\frac{12}{29}\right)\right) = \left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}\right) \)
The coordinates of the foot of the perpendicular M are \( \left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}\right) \).

(b) For the plane \( 3y + 4z = 6 \):
The given plane equation is \( 3y + 4z - 6 = 0 \), which can be written as \( 0x + 3y + 4z - 6 = 0 \).
The direction ratios of the normal to this plane are 0, 3, 4.
The equation of the line passing through the origin O(0, 0, 0) and parallel to this normal is:
\( \frac{x-0}{0}=\frac{y-0}{3}=\frac{z-0}{4} = \lambda \)
From this, we get \( x = 0, y = 3\lambda, z = 4\lambda \).
Let M be the foot of the perpendicular from the origin O to the plane. The coordinates of M are \( (0, 3\lambda, 4\lambda) \).
Since point M lies on the plane \( 3y + 4z - 6 = 0 \), its coordinates must satisfy the plane's equation:
\( 3(3\lambda) + 4(4\lambda) - 6 = 0 \)
\( 9\lambda + 16\lambda - 6 = 0 \)
\( 25\lambda - 6 = 0 \)
\( \implies \lambda = \frac{6}{25} \)
Now, substitute the value of \( \lambda \) back into the coordinates of M:
\( M = \left(0, 3\left(\frac{6}{25}\right), 4\left(\frac{6}{25}\right)\right) = \left(0, \frac{18}{25}, \frac{24}{25}\right) \)
The coordinates of the foot of the perpendicular M are \( \left(0, \frac{18}{25}, \frac{24}{25}\right) \).

(c) For the plane \( x + y + z = 1 \):
The given plane equation is \( x + y + z - 1 = 0 \).
The direction ratios of the normal to this plane are 1, 1, 1.
The equation of the line passing through the origin O(0, 0, 0) and parallel to this normal is:
\( \frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1} = \lambda \)
From this, we get \( x = \lambda, y = \lambda, z = \lambda \).
Let M be the foot of the perpendicular from the origin O to the plane. The coordinates of M are \( (\lambda, \lambda, \lambda) \).
Since point M lies on the plane \( x + y + z - 1 = 0 \), its coordinates must satisfy the plane's equation:
\( \lambda + \lambda + \lambda - 1 = 0 \)
\( 3\lambda - 1 = 0 \)
\( \implies \lambda = \frac{1}{3} \)
Now, substitute the value of \( \lambda \) back into the coordinates of M:
\( M = \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) \)
The coordinates of the foot of the perpendicular M are \( \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) \).

(d) For the plane \( 5y + 8 = 0 \):
The given plane equation is \( 5y + 8 = 0 \), which can be written as \( 0x + 5y + 0z + 8 = 0 \).
The direction ratios of the normal to this plane are 0, 5, 0.
The equation of the line passing through the origin O(0, 0, 0) and parallel to this normal is:
\( \frac{x-0}{0}=\frac{y-0}{5}=\frac{z-0}{0} = \lambda \)
From this, we get \( x = 0, y = 5\lambda, z = 0 \).
Let M be the foot of the perpendicular from the origin O to the plane. The coordinates of M are \( (0, 5\lambda, 0) \).
Since point M lies on the plane \( 5y + 8 = 0 \), its coordinates must satisfy the plane's equation:
\( 5(5\lambda) + 8 = 0 \)
\( 25\lambda + 8 = 0 \)
\( \implies \lambda = -\frac{8}{25} \)
Now, substitute the value of \( \lambda \) back into the coordinates of M:
\( M = \left(0, 5\left(-\frac{8}{25}\right), 0\right) = \left(0, -\frac{8}{5}, 0\right) \)
The coordinates of the foot of the perpendicular M are \( \left(0, -\frac{8}{5}, 0\right) \).
In simple words: To find the foot of the perpendicular from the origin to a plane, we first identify the direction ratios of the plane's normal vector. These ratios define a line passing through the origin and perpendicular to the plane. We then express the coordinates of any point on this line using a parameter \( \lambda \). By substituting these parametric coordinates into the plane's equation, we solve for \( \lambda \). Finally, we plug this \( \lambda \) value back into the parametric coordinates to get the exact location of the foot of the perpendicular.

Exam Tip: The foot of the perpendicular from the origin to a plane lies on the line passing through the origin and perpendicular to the plane. This line's direction ratios are the coefficients of x, y, z in the plane equation.

 

Question 5. નીચેના પૈકી પ્રત્યેક સમતલનાં સદિશ અને કાર્તેઝિય સમીકરણ શોધો :
(a) જે (1, 0, -2) માંથી પસાર થાય અને જેનો અભિલંબ સદિશ \( \hat{i} + \hat{j} – \hat{k} \) હોય.
(b) જે (1, 4, 6) માંથી પસાર થાય અને જેનો અભિલંબ સદિશ \( \hat{i} – 2\hat{j} + \hat{k} \) હોય.
Answer:
(a) For the plane passing through (1, 0, -2) and having normal vector \( \hat{i} + \hat{j} – \hat{k} \):
The position vector of the given point is \( \vec{a} = 1\hat{i} + 0\hat{j} - 2\hat{k} = \hat{i} - 2\hat{k} \).
The normal vector is \( \vec{n} = \hat{i} + \hat{j} – \hat{k} \).
The vector equation of a plane passing through a point with position vector \( \vec{a} \) and normal to vector \( \vec{n} \) is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \).
Substituting the given values:
\( (\vec{r} - (\hat{i} - 2\hat{k})) \cdot (\hat{i} + \hat{j} – \hat{k}) = 0 \)
This is the vector equation of the plane.
To find the Cartesian equation, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Substitute \( \vec{r} \) into the vector equation:
\( ((x\hat{i} + y\hat{j} + z\hat{k}) - (\hat{i} - 2\hat{k})) \cdot (\hat{i} + \hat{j} – \hat{k}) = 0 \)
\( ((x - 1)\hat{i} + y\hat{j} + (z + 2)\hat{k}) \cdot (\hat{i} + \hat{j} – \hat{k}) = 0 \)
Performing the dot product:
\( (x - 1)(1) + y(1) + (z + 2)(-1) = 0 \)
\( x - 1 + y - z - 2 = 0 \)
\( x + y - z = 3 \)
This is the Cartesian equation of the plane.
(b) For the plane passing through (1, 4, 6) and having normal vector \( \hat{i} – 2\hat{j} + \hat{k} \):
The position vector of the given point is \( \vec{a} = 1\hat{i} + 4\hat{j} + 6\hat{k} = \hat{i} + 4\hat{j} + 6\hat{k} \).
The normal vector is \( \vec{n} = \hat{i} – 2\hat{j} + \hat{k} \).
The vector equation of the plane is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \).
Substituting the given values:
\( (\vec{r} - (\hat{i} + 4\hat{j} + 6\hat{k})) \cdot (\hat{i} – 2\hat{j} + \hat{k}) = 0 \)
This is the vector equation of the plane.
To find the Cartesian equation, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Substitute \( \vec{r} \) into the vector equation:
\( ((x\hat{i} + y\hat{j} + z\hat{k}) - (\hat{i} + 4\hat{j} + 6\hat{k})) \cdot (\hat{i} – 2\hat{j} + \hat{k}) = 0 \)
\( ((x - 1)\hat{i} + (y - 4)\hat{j} + (z - 6)\hat{k}) \cdot (\hat{i} – 2\hat{j} + \hat{k}) = 0 \)
Performing the dot product:
\( (x - 1)(1) + (y - 4)(-2) + (z - 6)(1) = 0 \)
\( x - 1 - 2y + 8 + z - 6 = 0 \)
\( x - 2y + z + 1 = 0 \)
This is the Cartesian equation of the plane.
In simple words: For each part, we are given a point and a normal vector. First, we write the vector equation of the plane using the formula \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \), where \( \vec{a} \) is the position vector of the given point and \( \vec{n} \) is the normal vector. To get the Cartesian equation, we replace \( \vec{r} \) with \( x\hat{i} + y\hat{j} + z\hat{k} \) and perform the dot product, then simplify the expression.

Exam Tip: The vector equation of a plane passing through a point \( \vec{a} \) and having a normal vector \( \vec{n} \) is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \). Remember that \( \vec{a} \) is the position vector of the given point.

 

Question 6. નીચેના પૈકી આપેલ પ્રત્યેક પ્રશ્નમાં આપેલાં ત્રણ બિંદુઓમાંથી પસાર થતા સમતલનું સમીકરણ મેળવો ઃ
(a) (1, 1, -1), (6, 4, -5), (-4, -2, 3)
(b) (1, 1, 0), (1, 2, 1), (-2, 2, -1)
Answer:
(a) For the plane passing through (1, 1, -1), (6, 4, -5), (-4, -2, 3):
Let the three given points be \( A(x_1, y_1, z_1) = (1, 1, -1) \), \( B(x_2, y_2, z_2) = (6, 4, -5) \), and \( C(x_3, y_3, z_3) = (-4, -2, 3) \).
The equation of a plane passing through three non-collinear points is given by the determinant form:
\[ \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0 \]
Substitute the coordinates of the points:
\[ \begin{vmatrix} x-1 & y-1 & z-(-1) \\ 6-1 & 4-1 & -5-(-1) \\ -4-1 & -2-1 & 3-(-1) \end{vmatrix} = 0 \]
Simplify the entries in the determinant:
\[ \begin{vmatrix} x-1 & y-1 & z+1 \\ 5 & 3 & -4 \\ -5 & -3 & 4 \end{vmatrix} = 0 \]
Expand the determinant along the first row:
\( (x-1)((3)(4) - (-4)(-3)) - (y-1)((5)(4) - (-4)(-5)) + (z+1)((5)(-3) - (3)(-5)) = 0 \)
\( (x-1)(12 - 12) - (y-1)(20 - 20) + (z+1)(-15 + 15) = 0 \)
\( (x-1)(0) - (y-1)(0) + (z+1)(0) = 0 \)
\( 0 = 0 \)
This result indicates that the three given points A, B, and C are collinear.
When three points are collinear, an infinite number of planes can pass through them. Thus, there isn't a unique plane defined by these three points.
(b) For the plane passing through (1, 1, 0), (1, 2, 1), (-2, 2, -1):
Let the three given points be \( A(x_1, y_1, z_1) = (1, 1, 0) \), \( B(x_2, y_2, z_2) = (1, 2, 1) \), and \( C(x_3, y_3, z_3) = (-2, 2, -1) \).
The equation of a plane passing through three points is:
\[ \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0 \]
Substitute the coordinates of the points:
\[ \begin{vmatrix} x-1 & y-1 & z-0 \\ 1-1 & 2-1 & 1-0 \\ -2-1 & 2-1 & -1-0 \end{vmatrix} = 0 \]
Simplify the entries in the determinant:
\[ \begin{vmatrix} x-1 & y-1 & z \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = 0 \]
Expand the determinant along the first row:
\( (x-1)((1)(-1) - (1)(1)) - (y-1)((0)(-1) - (1)(-3)) + z((0)(1) - (1)(-3)) = 0 \)
\( (x-1)(-1 - 1) - (y-1)(0 + 3) + z(0 + 3) = 0 \)
\( (x-1)(-2) - (y-1)(3) + z(3) = 0 \)
\( -2x + 2 - 3y + 3 + 3z = 0 \)
\( -2x - 3y + 3z + 5 = 0 \)
Multiplying by -1 to make the coefficient of x positive:
\( 2x + 3y - 3z - 5 = 0 \)
So, \( 2x + 3y - 3z = 5 \)
This is the required equation of the plane.
In simple words: To find the equation of a plane that passes through three given points, we use a determinant. We set up a 3x3 determinant where the first row contains \( (x-x_1), (y-y_1), (z-z_1) \), and the next two rows are formed by subtracting the coordinates of the first point from the other two points. If the determinant equals zero, the points are collinear, and many planes can pass through them. Otherwise, expanding the determinant gives the Cartesian equation of the unique plane.

Exam Tip: When finding the equation of a plane through three points, ensure the points are not collinear. If they are, the determinant will be zero, indicating an infinite number of planes.

 

Question 7. સમતલ \( 2x + y – z = 5 \) દ્વારા અક્ષો પર કપાતા અંતઃખંડ શોધો.
Answer:
The given equation of the plane is \( 2x + y – z = 5 \).
To find the intercepts cut off by the plane on the axes, we need to convert the equation into the intercept form, which is \( \frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1 \).
Divide the entire equation by 5:
\( \frac{2x}{5} + \frac{y}{5} - \frac{z}{5} = \frac{5}{5} \)
\( \frac{2x}{5} + \frac{y}{5} - \frac{z}{5} = 1 \)
To match the standard intercept form, we rewrite the first term:
\( \frac{x}{5/2} + \frac{y}{5} + \frac{z}{-5} = 1 \)
Comparing this with \( \frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1 \), we get the intercepts:
X-intercept, \( A = \frac{5}{2} \)
Y-intercept, \( B = 5 \)
Z-intercept, \( C = -5 \)
Thus, the intercepts cut off by the plane on the axes are \( \frac{5}{2}, 5 \), and \( -5 \) respectively.
In simple words: To find where a plane crosses the X, Y, and Z axes, we simply rewrite its equation. We divide all parts of the equation by the constant term on the right side so that the right side becomes 1. Then, the numbers under \( x, y, \) and \( z \) are the points where the plane cuts each axis.

Exam Tip: To find intercepts, always rearrange the plane equation into the form \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \). The constants 'a', 'b', and 'c' will represent the intercepts on the x, y, and z axes, respectively.

 

Question 8. Y–અક્ષ પર 3 અંતઃખંડવાળા અને ZOX સમતલને સમાંતર સમતલનું સમીકરણ શોધો.
Answer:
A plane parallel to the ZOX plane (which is the xz-plane) has an equation of the form \( y = b \), where \( b \) is a constant.
In this equation, \( b \) represents the Y-intercept of the plane.
We are given that the plane has an intercept of 3 on the Y-axis, so \( b = 3 \).
Therefore, the equation of the required plane is \( y = 3 \).
In simple words: A plane that runs parallel to the ZOX plane always has an equation like \( y = \text{a number} \). This number tells us where it crosses the Y-axis. Since we're told it crosses the Y-axis at 3, its equation is simply \( y = 3 \).

Exam Tip: Understand that a plane parallel to a coordinate plane (like ZOX) will have an equation where only the coordinate corresponding to the perpendicular axis appears. For the ZOX plane, the normal is along the Y-axis, so the equation is \( y = \text{constant} \).

 

Question 9. સમતલો \( 3x – y + 2z – 4 = 0 \) અને \( x + y + z − 2 = 0 \) ના છેદમાંથી તથા બિંદુ (2, 2, 1) માંથી પસાર થતા સમતલનું સમીકરણ શોધો.
Answer:
Let the two given planes be \( P_1: 3x – y + 2z – 4 = 0 \) and \( P_2: x + y + z – 2 = 0 \).
The equation of a plane passing through the intersection of \( P_1 \) and \( P_2 \) is given by the formula \( P_1 + \lambda P_2 = 0 \), where \( \lambda \) is an arbitrary constant.
So, the equation of the required plane is:
\( (3x – y + 2z – 4) + \lambda(x + y + z – 2) = 0 \) (i)
We are given that this plane passes through the point (2, 2, 1). To find \( \lambda \), substitute the coordinates of this point into equation (i):
\( (3(2) – 2 + 2(1) – 4) + \lambda(2 + 2 + 1 – 2) = 0 \)
\( (6 – 2 + 2 – 4) + \lambda(3) = 0 \)
\( (2) + 3\lambda = 0 \)
\( \implies 3\lambda = -2 \)
\( \implies \lambda = -\frac{2}{3} \)
Now, substitute the value of \( \lambda \) back into equation (i):
\( (3x – y + 2z – 4) - \frac{2}{3}(x + y + z – 2) = 0 \)
To clear the fraction, multiply the entire equation by 3:
\( 3(3x – y + 2z – 4) - 2(x + y + z – 2) = 0 \)
\( 9x – 3y + 6z – 12 – 2x – 2y – 2z + 4 = 0 \)
Combine the like terms:
\( (9x - 2x) + (-3y - 2y) + (6z - 2z) + (-12 + 4) = 0 \)
\( 7x - 5y + 4z - 8 = 0 \)
This is the required Cartesian equation of the plane.
In simple words: We are looking for a plane that goes through the line where two other planes cross. We write a general equation for such a plane using a special number called lambda. Then, we use the fact that this new plane also passes through a specific point. By putting the coordinates of that point into our general equation, we can figure out the value of lambda. Finally, we plug lambda back into the general equation and simplify it to get the plane's exact equation.

Exam Tip: The form \( P_1 + \lambda P_2 = 0 \) is crucial for planes passing through the intersection of two given planes. This method is efficient for solving such problems.

 

Question 10. સમતલો \( \vec{r} \cdot (2\hat{i} + 2\hat{j} – 3\hat{k}) = 7 \) અને \( \vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) = 9 \) ના છેદમાંથી તથા બિંદુ (2, 1, 3) માંથી પસાર થતા સમતલનું સદિશ સમીકરણ શોધો.
Answer:
Let the two given planes be \( P_1: \vec{r} \cdot (2\hat{i} + 2\hat{j} – 3\hat{k}) - 7 = 0 \) and \( P_2: \vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 9 = 0 \).
The equation of a plane passing through the intersection of \( P_1 \) and \( P_2 \) is \( P_1 + \lambda P_2 = 0 \).
So, the equation of the required plane is:
\( [\vec{r} \cdot (2\hat{i} + 2\hat{j} – 3\hat{k}) - 7] + \lambda [\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 9] = 0 \) (i)
The plane passes through the point (2, 1, 3), which means \( \vec{r} = 2\hat{i} + 1\hat{j} + 3\hat{k} \).
Substitute this position vector into equation (i) to find \( \lambda \):
\( [(2\hat{i} + \hat{j} + 3\hat{k}) \cdot (2\hat{i} + 2\hat{j} – 3\hat{k}) - 7] + \lambda [(2\hat{i} + \hat{j} + 3\hat{k}) \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 9] = 0 \)
Calculate the dot products:
\( [((2)(2) + (1)(2) + (3)(-3)) - 7] + \lambda [((2)(2) + (1)(5) + (3)(3)) - 9] = 0 \)
\( [(4 + 2 - 9) - 7] + \lambda [(4 + 5 + 9) - 9] = 0 \)
\( [-3 - 7] + \lambda [9] = 0 \)
\( -10 + 9\lambda = 0 \)
\( \implies 9\lambda = 10 \)
\( \implies \lambda = \frac{10}{9} \)
Substitute the value of \( \lambda \) back into equation (i):
\( [\vec{r} \cdot (2\hat{i} + 2\hat{j} – 3\hat{k}) - 7] + \frac{10}{9} [\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 9] = 0 \)
Multiply the entire equation by 9 to remove the fraction:
\( 9[\vec{r} \cdot (2\hat{i} + 2\hat{j} – 3\hat{k}) - 7] + 10[\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 9] = 0 \)
\( 9\vec{r} \cdot (2\hat{i} + 2\hat{j} – 3\hat{k}) - 63 + 10\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 90 = 0 \)
Rearrange and combine the \( \vec{r} \) terms:
\( \vec{r} \cdot [9(2\hat{i} + 2\hat{j} – 3\hat{k}) + 10(2\hat{i} + 5\hat{j} + 3\hat{k})] - 63 - 90 = 0 \)
\( \vec{r} \cdot [(18\hat{i} + 18\hat{j} – 27\hat{k}) + (20\hat{i} + 50\hat{j} + 30\hat{k})] - 153 = 0 \)
\( \vec{r} \cdot [(18+20)\hat{i} + (18+50)\hat{j} + (-27+30)\hat{k}] - 153 = 0 \)
\( \vec{r} \cdot (38\hat{i} + 68\hat{j} + 3\hat{k}) - 153 = 0 \)
So, the final vector equation of the plane is \( \vec{r} \cdot (38\hat{i} + 68\hat{j} + 3\hat{k}) = 153 \).
In simple words: We start with the vector equations of two planes and combine them using a scalar \( \lambda \) to form a new equation that represents any plane passing through their intersection. Then, we use the given point that the desired plane passes through. By plugging this point's position vector into our combined equation, we solve for \( \lambda \). Finally, we substitute \( \lambda \) back into the equation and simplify it to get the exact vector equation of the plane we need.

Exam Tip: When combining \( \vec{r} \) terms in vector equations, distribute the scalar multipliers and then group the corresponding \( \hat{i}, \hat{j}, \hat{k} \) components.

 

Question 11. સમતલો \( x + y + z = 1 \) x \( 2x + 3y + 4z = 5 \) ની છેદરેખામાંથી પસાર થતા તથા સમતલ \( x − y + z = 0 \) ને લંબ હોય તેવા સમતલનું સમીકરણ શોધો.
Answer:
Let the two given planes be \( P_1: x + y + z - 1 = 0 \) and \( P_2: 2x + 3y + 4z - 5 = 0 \).
The equation of a plane passing through the intersection of \( P_1 \) and \( P_2 \) is given by \( P_1 + \lambda P_2 = 0 \):
\( (x + y + z - 1) + \lambda(2x + 3y + 4z - 5) = 0 \)
Rearranging the terms, we get the equation in the form \( Ax + By + Cz + D = 0 \):
\( (1 + 2\lambda)x + (1 + 3\lambda)y + (1 + 4\lambda)z - (1 + 5\lambda) = 0 \) (i)
The normal vector to this plane is \( \vec{n_1} = (1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k} \).
We are also given that the required plane is perpendicular to the plane \( P_3: x - y + z = 0 \).
The normal vector to plane \( P_3 \) is \( \vec{n_2} = 1\hat{i} - 1\hat{j} + 1\hat{k} = \hat{i} - \hat{j} + \hat{k} \).
Since the two planes are perpendicular, their normal vectors must also be perpendicular. This means their dot product is zero:
\( \vec{n_1} \cdot \vec{n_2} = 0 \)
\( ((1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = 0 \)
\( (1 + 2\lambda)(1) + (1 + 3\lambda)(-1) + (1 + 4\lambda)(1) = 0 \)
\( 1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0 \)
\( (2\lambda - 3\lambda + 4\lambda) + (1 - 1 + 1) = 0 \)
\( 3\lambda + 1 = 0 \)
\( \implies 3\lambda = -1 \)
\( \implies \lambda = -\frac{1}{3} \)
Now, substitute the value of \( \lambda \) back into equation (i):
\( (1 + 2(-\frac{1}{3}))x + (1 + 3(-\frac{1}{3}))y + (1 + 4(-\frac{1}{3}))z - (1 + 5(-\frac{1}{3})) = 0 \)
\( (1 - \frac{2}{3})x + (1 - 1)y + (1 - \frac{4}{3})z - (1 - \frac{5}{3}) = 0 \)
\( \left(\frac{3-2}{3}\right)x + (0)y + \left(\frac{3-4}{3}\right)z - \left(\frac{3-5}{3}\right) = 0 \)
\( \frac{1}{3}x + 0y - \frac{1}{3}z - \left(-\frac{2}{3}\right) = 0 \)
\( \frac{1}{3}x - \frac{1}{3}z + \frac{2}{3} = 0 \)
Multiply the entire equation by 3 to simplify:
\( x - z + 2 = 0 \)
This is the required equation of the plane.
In simple words: We want to find a plane that goes through the meeting point of two other planes and is also perpendicular to a third plane. First, we write a general equation for planes passing through the intersection of the first two using a variable \( \lambda \). Then, we find the normal vector for this general plane. We also get the normal vector of the third plane. Because the general plane is perpendicular to the third plane, the dot product of their normal vectors must be zero. This lets us solve for \( \lambda \). Finally, we put the value of \( \lambda \) back into the general equation and simplify it to get the final plane equation.

Exam Tip: When two planes are perpendicular, their normal vectors are also perpendicular. This means their dot product is zero. This property is key to finding the unknown parameter.

 

Question 12. સમતલના સદિશ સમીકરણ \( \vec{r} \cdot (2\hat{i} + 2\hat{j} – 3\hat{k}) = 5 \) અને \( \vec{r} \cdot (3\hat{i} – 3\hat{j} + 5\hat{k}) = 3 \) છે. તેમની વચ્ચેનો ખૂણો શોધો.
Answer:
The vector equations of the two planes are:
Plane \( P_1: \vec{r} \cdot (2\hat{i} + 2\hat{j} – 3\hat{k}) = 5 \)
Plane \( P_2: \vec{r} \cdot (3\hat{i} – 3\hat{j} + 5\hat{k}) = 3 \)
The normal vector to plane \( P_1 \) is \( \vec{n_1} = 2\hat{i} + 2\hat{j} – 3\hat{k} \).
The normal vector to plane \( P_2 \) is \( \vec{n_2} = 3\hat{i} – 3\hat{j} + 5\hat{k} \).
First, calculate the magnitudes of the normal vectors:
\( |\vec{n_1}| = \sqrt{2^2 + 2^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \)
\( |\vec{n_2}| = \sqrt{3^2 + (-3)^2 + 5^2} = \sqrt{9 + 9 + 25} = \sqrt{43} \)
Next, calculate the dot product of the normal vectors:
\( \vec{n_1} \cdot \vec{n_2} = (2\hat{i} + 2\hat{j} – 3\hat{k}) \cdot (3\hat{i} – 3\hat{j} + 5\hat{k}) \)
\( = (2)(3) + (2)(-3) + (-3)(5) \)
\( = 6 - 6 - 15 = -15 \)
Let \( \theta \) be the angle between the two planes. The cosine of the angle between two planes is given by:
\( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} \)
Substitute the calculated values into the formula:
\( \cos \theta = \frac{|-15|}{(\sqrt{17})(\sqrt{43})} = \frac{15}{\sqrt{17 \times 43}} = \frac{15}{\sqrt{731}} \)
Therefore, the angle between the planes is \( \theta = \cos^{-1}\left(\frac{15}{\sqrt{731}}\right) \).
In simple words: To find the angle between two planes, we first identify their normal vectors. Then, we calculate the length of each normal vector and their dot product. Finally, we use the formula \( \cos \theta = \frac{|\text{dot product of normals}|}{\text{product of their lengths}} \) to find the cosine of the angle, and then take the inverse cosine to get the angle itself.

Exam Tip: The angle between two planes is defined as the angle between their normal vectors. Always use the absolute value of the dot product in the numerator to ensure the angle found is acute.

 

Question 13. નીચેના પૈકી પ્રત્યેક પ્રશ્નમાં આપેલા સમતલ સમાંતર છે કે પરસ્પર લંબ છે તે નક્કી કરો અને જો આ પૈકી એક પણ ન હોય, તો તેમની વચ્ચેનો ખૂણો શોધો :
(a) \( 7x + 5y + 6z + 30 = 0 \) અને \( 3x – y – 10z + 4 = 0 \)
(b) \( 2x + y + 3z- 2 = 0 \) અને \( x – 2y + 5 = 0 \)
(c) \( 2x-2y+4z + 5 = 0 \) અને \( 3x-3y+ 6z – 1 = 0 \)
(d) \( 2x – y + 3z − 1 = 0 \) અને \( 2x – y + 3z + 3 = 0 \)
(e) \( 4x+8y + z-8 = 0 \) અને \( y + z − 4 = 0 \)
Answer:
(a) For the planes \( 7x + 5y + 6z + 30 = 0 \) and \( 3x – y – 10z + 4 = 0 \):
The normal vector for the first plane is \( \vec{n_1} = 7\hat{i} + 5\hat{j} + 6\hat{k} \).
The normal vector for the second plane is \( \vec{n_2} = 3\hat{i} - \hat{j} - 10\hat{k} \).
Check for parallelism:
The ratios of the corresponding components are \( \frac{7}{3} \), \( \frac{5}{-1} = -5 \), and \( \frac{6}{-10} = -\frac{3}{5} \).
Since \( \frac{7}{3} \neq -5 \), the planes are not parallel.
Check for perpendicularity:
Calculate the dot product of their normal vectors:
\( \vec{n_1} \cdot \vec{n_2} = (7)(3) + (5)(-1) + (6)(-10) \)
\( = 21 - 5 - 60 = -44 \)
Since \( \vec{n_1} \cdot \vec{n_2} \neq 0 \), the planes are not perpendicular.
Calculate the angle between them:
\( |\vec{n_1}| = \sqrt{7^2 + 5^2 + 6^2} = \sqrt{49 + 25 + 36} = \sqrt{110} \)
\( |\vec{n_2}| = \sqrt{3^2 + (-1)^2 + (-10)^2} = \sqrt{9 + 1 + 100} = \sqrt{110} \)
Let \( \theta \) be the angle between the planes.
\( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|-44|}{\sqrt{110}\sqrt{110}} = \frac{44}{110} = \frac{2}{5} \)
Therefore, \( \theta = \cos^{-1}\left(\frac{2}{5}\right) \).
(b) For the planes \( 2x + y + 3z- 2 = 0 \) and \( x – 2y + 5 = 0 \):
The normal vector for the first plane is \( \vec{n_1} = 2\hat{i} + \hat{j} + 3\hat{k} \).
The normal vector for the second plane is \( \vec{n_2} = 1\hat{i} - 2\hat{j} + 0\hat{k} = \hat{i} - 2\hat{j} \).
Check for parallelism:
The ratios of the corresponding components are \( \frac{2}{1} = 2 \) and \( \frac{1}{-2} \). Since these are not equal, the planes are not parallel.
Check for perpendicularity:
Calculate the dot product of their normal vectors:
\( \vec{n_1} \cdot \vec{n_2} = (2)(1) + (1)(-2) + (3)(0) \)
\( = 2 - 2 + 0 = 0 \)
Since \( \vec{n_1} \cdot \vec{n_2} = 0 \), the planes are perpendicular to each other.
(c) For the planes \( 2x-2y+4z + 5 = 0 \) and \( 3x-3y+ 6z – 1 = 0 \):
The normal vector for the first plane is \( \vec{n_1} = 2\hat{i} - 2\hat{j} + 4\hat{k} \).
The normal vector for the second plane is \( \vec{n_2} = 3\hat{i} - 3\hat{j} + 6\hat{k} \).
Check for parallelism:
Compare the ratios of the corresponding components:
\( \frac{a_1}{a_2} = \frac{2}{3} \)
\( \frac{b_1}{b_2} = \frac{-2}{-3} = \frac{2}{3} \)
\( \frac{c_1}{c_2} = \frac{4}{6} = \frac{2}{3} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{2}{3} \), the normal vectors are parallel. Therefore, the planes are parallel.
(d) For the planes \( 2x – y + 3z − 1 = 0 \) and \( 2x – y + 3z + 3 = 0 \):
The normal vector for the first plane is \( \vec{n_1} = 2\hat{i} - \hat{j} + 3\hat{k} \).
The normal vector for the second plane is \( \vec{n_2} = 2\hat{i} - \hat{j} + 3\hat{k} \).
Check for parallelism:
Compare the ratios of the corresponding components:
\( \frac{a_1}{a_2} = \frac{2}{2} = 1 \)
\( \frac{b_1}{b_2} = \frac{-1}{-1} = 1 \)
\( \frac{c_1}{c_2} = \frac{3}{3} = 1 \)
Since the ratios of the corresponding components are equal (in fact, the normal vectors are identical), the planes are parallel.
(e) For the planes \( 4x+8y + z-8 = 0 \) and \( y + z − 4 = 0 \):
The normal vector for the first plane is \( \vec{n_1} = 4\hat{i} + 8\hat{j} + 1\hat{k} \).
The normal vector for the second plane is \( \vec{n_2} = 0\hat{i} + 1\hat{j} + 1\hat{k} \).
Check for parallelism:
The ratios of the corresponding components are \( \frac{4}{0} \) (undefined), \( \frac{8}{1} = 8 \), and \( \frac{1}{1} = 1 \). These are not equal, so the planes are not parallel.
Check for perpendicularity:
Calculate the dot product of their normal vectors:
\( \vec{n_1} \cdot \vec{n_2} = (4)(0) + (8)(1) + (1)(1) \)
\( = 0 + 8 + 1 = 9 \)
Since \( \vec{n_1} \cdot \vec{n_2} \neq 0 \), the planes are not perpendicular.
Calculate the angle between them:
\( |\vec{n_1}| = \sqrt{4^2 + 8^2 + 1^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9 \)
\( |\vec{n_2}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \)
Let \( \theta \) be the angle between the planes.
\( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|9|}{(9)(\sqrt{2})} = \frac{1}{\sqrt{2}} \)
Therefore, \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \).
In simple words: For each pair of planes, we first find their normal vectors. Then, we check for parallelism by comparing if the components of these normal vectors are proportional. If they are, the planes are parallel. Next, we check for perpendicularity by calculating the dot product of the normal vectors. If the dot product is zero, the planes are perpendicular. If the planes are neither parallel nor perpendicular, we calculate the angle between them using the formula \( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} \), where \( \vec{n_1} \) and \( \vec{n_2} \) are the normal vectors.

Exam Tip: Always check for parallelism and perpendicularity first, as they are simpler conditions. If neither applies, then calculate the angle using the dot product formula.

 

Question 13. Determine if each of the given planes is parallel or perpendicular, and if neither, find the angle between them:
(d) \( 2x - y + 3z - 1 = 0 \) and \( 2x - y + 3z + 3 = 0 \)
Answer:
The first plane \( \pi_1 \) is given by: \( 2x - y + 3z - 1 = 0 \)
Its normal vector is \( \overrightarrow{n_1} = 2\hat{i} - \hat{j} + 3\hat{k} \).
The second plane \( \pi_2 \) is given by: \( 2x - y + 3z + 3 = 0 \)
Its normal vector is \( \overrightarrow{n_2} = 2\hat{i} - \hat{j} + 3\hat{k} \).
We can clearly see that \( \overrightarrow{n_1} \) is equal to \( \overrightarrow{n_2} \).
Since their normal vectors are parallel (actually identical in this case), the given planes are also parallel to each other.
In simple words: When the normal vectors of two planes are the same or parallel, it means the planes themselves are also parallel and will never cross each other.

Exam Tip: To check if two planes are parallel, compare their normal vectors. If the normal vectors are scalar multiples of each other, the planes are parallel.

 

Question 13. Determine if each of the given planes is parallel or perpendicular, and if neither, find the angle between them:
(e) \( 4x + 8y + z - 8 = 0 \) and \( y + z - 4 = 0 \)
Answer:
The first plane \( \pi_1 \) is given by: \( 4x + 8y + z - 8 = 0 \)
Its normal vector is \( \overrightarrow{n_1} = 4\hat{i} + 8\hat{j} + \hat{k} \).
The second plane \( \pi_2 \) is given by: \( y + z - 4 = 0 \)
Its normal vector is \( \overrightarrow{n_2} = 0\hat{i} + \hat{j} + \hat{k} \).
First, calculate the dot product of the normal vectors:
\( \overrightarrow{n_1} \cdot \overrightarrow{n_2} = (4\hat{i} + 8\hat{j} + \hat{k}) \cdot (\hat{j} + \hat{k}) \)
\( = (4)(0) + (8)(1) + (1)(1) \)
\( = 0 + 8 + 1 \)
\( = 9 \)
Next, calculate the magnitudes of the normal vectors:
\( |\overrightarrow{n_1}| = \sqrt{(4)^2 + (8)^2 + (1)^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9 \)
\( |\overrightarrow{n_2}| = \sqrt{(0)^2 + (1)^2 + (1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \)
Since the dot product is not zero, the planes are not perpendicular. Also, since \( \overrightarrow{n_1} \) is not a scalar multiple of \( \overrightarrow{n_2} \), the planes are not parallel.
We will now find the angle \( \theta \) between the planes using the formula:
\( \cos \theta = \frac{|\overrightarrow{n_1} \cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}| |\overrightarrow{n_2}|} \)
\( = \frac{|9|}{(9)(\sqrt{2})} \)
\( = \frac{9}{9\sqrt{2}} \)
\( = \frac{1}{\sqrt{2}} \)
Now, find \( \theta \):
\( \cos \theta = \frac{1}{\sqrt{2}} \)
\( \implies \theta = 45^\circ \)
Thus, the angle between the given planes is \( 45^\circ \).
In simple words: To find the angle between two planes, we first get their normal vectors. Then, we find the dot product of these vectors and divide it by the product of their lengths. This gives us the cosine of the angle. From that, we can figure out the actual angle.

Exam Tip: Remember that the angle between two planes is defined as the angle between their normal vectors. If the dot product of the normal vectors is zero, the planes are perpendicular. If the normal vectors are parallel, the planes are parallel.

 

Question 14. Find the distance of each of the given points from the corresponding given plane:

Answer:
(a) For the point (0, 0, 0) and plane \( 3x - 4y + 12z = 3 \):
The given plane's equation is \( 3x - 4y + 12z - 3 = 0 \).
The perpendicular distance from a point \( (x_1, y_1, z_1) \) to a plane \( ax + by + cz + d = 0 \) is given by:
\( d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \)
Substituting the values \( (x_1, y_1, z_1) = (0, 0, 0) \) and \( a=3, b=-4, c=12, d=-3 \):
\( d = \frac{|3(0) - 4(0) + 12(0) - 3|}{\sqrt{3^2 + (-4)^2 + 12^2}} \)
\( = \frac{|-3|}{\sqrt{9 + 16 + 144}} \)
\( = \frac{3}{\sqrt{169}} \)
\( = \frac{3}{13} \)
The distance is \( \frac{3}{13} \) units.
In simple words: We used a formula to find how far the point (0, 0, 0) is from the plane. We put the point's numbers and the plane's numbers into the formula, and after doing the math, we found the distance.

Exam Tip: Remember to rearrange the plane equation to the standard form \( ax + by + cz + d = 0 \) before applying the distance formula, ensuring the constant term is on the same side as \( x, y, z \).

 

(b) For the point (3, -2, 1) and plane \( 2x - y + 2z + 3 = 0 \):
The given plane's equation is \( 2x - y + 2z + 3 = 0 \).
Using the distance formula \( d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \):
Substituting the values \( (x_1, y_1, z_1) = (3, -2, 1) \) and \( a=2, b=-1, c=2, d=3 \):
\( d = \frac{|2(3) - (-2) + 2(1) + 3|}{\sqrt{2^2 + (-1)^2 + 2^2}} \)
\( = \frac{|6 + 2 + 2 + 3|}{\sqrt{4 + 1 + 4}} \)
\( = \frac{|13|}{\sqrt{9}} \)
\( = \frac{13}{3} \)
The distance is \( \frac{13}{3} \) units.
In simple words: For the second point and its plane, we put their specific numbers into the distance formula. After doing the calculations carefully, we got a distance of thirteen-thirds.

Exam Tip: Always be careful with negative signs when substituting coordinates into the distance formula, especially for terms like \( -y_1 \) or \( -d \).

 

(c) For the point (2, 3, -5) and plane \( x + 2y - 2z = 9 \):
The given plane's equation is \( x + 2y - 2z - 9 = 0 \).
Using the distance formula \( d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \):
Substituting the values \( (x_1, y_1, z_1) = (2, 3, -5) \) and \( a=1, b=2, c=-2, d=-9 \):
\( d = \frac{|1(2) + 2(3) - 2(-5) - 9|}{\sqrt{1^2 + 2^2 + (-2)^2}} \)
\( = \frac{|2 + 6 + 10 - 9|}{\sqrt{1 + 4 + 4}} \)
\( = \frac{|9|}{\sqrt{9}} \)
\( = \frac{9}{3} \)
\( = 3 \)
The distance is 3 units.
In simple words: For the third point and its plane, we used the distance formula. We added the absolute values of the terms in the numerator and divided by the square root of the sum of squares of coefficients. This calculation gave us a distance of 3.

Exam Tip: Remember that the distance between a point and a plane must always be a non-negative value. The absolute value in the numerator ensures this.

 

(d) For the point (-6, 0, 0) and plane \( 2x - 3y + 6z - 2 = 0 \):
The given plane's equation is \( 2x - 3y + 6z - 2 = 0 \).
Using the distance formula \( d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \):
Substituting the values \( (x_1, y_1, z_1) = (-6, 0, 0) \) and \( a=2, b=-3, c=6, d=-2 \):
\( d = \frac{|2(-6) - 3(0) + 6(0) - 2|}{\sqrt{2^2 + (-3)^2 + 6^2}} \)
\( = \frac{|-12 - 0 + 0 - 2|}{\sqrt{4 + 9 + 36}} \)
\( = \frac{|-14|}{\sqrt{49}} \)
\( = \frac{14}{7} \)
\( = 2 \)
The distance is 2 units.
In simple words: For the last point and plane, we placed all the numbers into the distance formula. We calculated the top part (numerator) and the bottom part (denominator) and then divided them to get the final distance of 2.

Exam Tip: A common mistake is forgetting to take the absolute value of the numerator in the distance formula, which can lead to a negative "distance" if the point is on one side of the plane.