RBSE Solutions Class 12 Maths Chapter 14 Three Dimensional Geometry Exercise 14.5

Question 1. Find the shortest distance between the lines
\( \vec { r } = (i+2j+k)+\lambda (i-j+k) \)
and
\( \vec { r } = 2i-j-k+\mu (2i+j+2k) \)
Answer: First, we compare the given line equations with the general vector form \( \vec { r } = \vec { a } + \lambda \vec { b } \).
For the first line:
\( \vec { a_1 } = i+2j+k \)
\( \vec { b_1 } = i-j+k \)
For the second line:
\( \vec { a_2 } = 2i-j-k \)
\( \vec { b_2 } = 2i+j+2k \)
Next, we find the difference vector \( \vec { a_2 } - \vec { a_1 } \):
\( \vec { a_2 } - \vec { a_1 } = (2i-j-k) - (i+2j+k) \)
\( = (2-1)i + (-1-2)j + (-1-1)k \)
\( = i-3j-2k \)
Then, we calculate the cross product of the direction vectors \( \vec { b_1 } \times \vec { b_2 } \):
\( \vec { b_1 } \times \vec { b_2 } = (i-j+k) \times (2i+j+2k) \)
This can be found using a determinant:
\[ \begin{vmatrix} i & j & k \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} \] \( = i((-1)(2) - (1)(1)) - j((1)(2) - (1)(2)) + k((1)(1) - (-1)(2)) \)
\( = i(-2-1) - j(2-2) + k(1+2) \)
\( = -3i - 0j + 3k \)
\( = -3i+3k \)
Now, we find the magnitude of the cross product \( | \vec { b_1 } \times \vec { b_2 } | \):
\( | \vec { b_1 } \times \vec { b_2 } | = \sqrt{(-3)^2 + (3)^2} \)
\( = \sqrt{9+9} \)
\( = \sqrt{18} \)
\( = 3\sqrt{2} \)
Finally, the shortest distance \( d \) is given by the formula:
\( d = \left| \frac { (\vec { a_2 } - \vec { a_1 }) \cdot (\vec { b_1 } \times \vec { b_2 }) }{ | \vec { b_1 } \times \vec { b_2 } | } \right| \)
First, calculate the dot product \( (\vec { a_2 } - \vec { a_1 }) \cdot (\vec { b_1 } \times \vec { b_2 }) \):
\( (i-3j-2k) \cdot (-3i+3k) \)
\( = (1)(-3) + (-3)(0) + (-2)(3) \)
\( = -3 + 0 - 6 \)
\( = -9 \)
Substitute this into the distance formula:
\( d = \left| \frac { -9 }{ 3\sqrt{2} } \right| \)
\( d = \left| \frac { -3 }{ \sqrt{2} } \right| \)
\( d = \frac { 3 }{ \sqrt{2} } \)
To rationalize the denominator, multiply by \( \frac { \sqrt{2} }{ \sqrt{2} } \):
\( d = \frac { 3\sqrt{2} }{ 2 } \)
The shortest distance between the lines is \( \frac{3\sqrt{2}}{2} \). The shortest distance is always a positive value, representing a physical length.
In simple words: To find the shortest distance, first we identify the starting points and directions of both lines. Then we find the difference between the starting points and the cross product of the direction vectors. Finally, we use a formula involving the dot product and magnitude to get the smallest distance between them.

🎯 Exam Tip: Remember to identify \( \vec{a_1}, \vec{b_1}, \vec{a_2}, \vec{b_2} \) correctly from the given equations, as a small error here can propagate through the entire calculation. Always keep the absolute value for the final distance.

Question 2. Find the shortest distance between the lines
\( \frac { x+1 }{ 7 } = \frac { y+1 }{ -6 } = \frac { z+1 }{ 1 } \)
and
\( \frac { x-3 }{ 1 } = \frac { y-5 }{ -2 } = \frac { z-7 }{ 1 } \)
Answer: We compare the given Cartesian line equations with the general form \( \frac { x-x_1 }{ a_1 } = \frac { y-y_1 }{ b_1 } = \frac { z-z_1 }{ c_1 } \) and \( \frac { x-x_2 }{ a_2 } = \frac { y-y_2 }{ b_2 } = \frac { z-z_2 }{ c_2 } \).
For the first line:
\( x_1 = -1, y_1 = -1, z_1 = -1 \)
\( a_1 = 7, b_1 = -6, c_1 = 1 \)
For the second line:
\( x_2 = 3, y_2 = 5, z_2 = 7 \)
\( a_2 = 1, b_2 = -2, c_2 = 1 \)
The shortest distance \( d \) between two skew lines in Cartesian form is given by the formula:
\[ d = \left| \frac { \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} }{ \sqrt{(b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2 + (a_1b_2-a_2b_1)^2} } \right| \] First, calculate the numerator determinant:
\( x_2-x_1 = 3 - (-1) = 4 \)
\( y_2-y_1 = 5 - (-1) = 6 \)
\( z_2-z_1 = 7 - (-1) = 8 \)
So the determinant is:
\[ \begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} \] \( = 4((-6)(1) - (1)(-2)) - 6((7)(1) - (1)(1)) + 8((7)(-2) - (-6)(1)) \)
\( = 4(-6+2) - 6(7-1) + 8(-14+6) \)
\( = 4(-4) - 6(6) + 8(-8) \)
\( = -16 - 36 - 64 \)
\( = -116 \)
Next, calculate the terms for the denominator:
\( (b_1c_2-b_2c_1) = ((-6)(1) - (-2)(1)) = (-6+2) = -4 \)
\( (c_1a_2-c_2a_1) = ((1)(1) - (1)(7)) = (1-7) = -6 \)
\( (a_1b_2-a_2b_1) = ((7)(-2) - (1)(-6)) = (-14+6) = -8 \)
Now, find the denominator's square root:
\( \sqrt{(-4)^2 + (-6)^2 + (-8)^2} \)
\( = \sqrt{16+36+64} \)
\( = \sqrt{116} \)
Finally, substitute these values into the shortest distance formula:
\( d = \left| \frac { -116 }{ \sqrt{116} } \right| \)
\( d = \frac { 116 }{ \sqrt{116} } \)
\( d = \sqrt{116} \)
We can simplify \( \sqrt{116} \) as \( \sqrt{4 \times 29} = 2\sqrt{29} \).
The shortest distance between the lines is \( 2\sqrt{29} \). Understanding both vector and Cartesian forms helps in solving problems like this efficiently.
In simple words: We list the points and direction ratios for both lines. We then calculate a special number (a determinant) from these values for the top part of a fraction. For the bottom part, we find the square root of squared differences. Dividing the top by the bottom gives us the shortest distance between the two lines.

🎯 Exam Tip: Be very careful with the signs when calculating the determinant and the terms in the denominator. A common mistake is a sign error during subtraction or multiplication.

Question 3. Find the shortest distance between the lines
\( \vec { r } = (i+2j+3k)+\lambda (i-3j+2k) \)
and
\( \vec { r } = (4i+5j+6k)+\mu (2i+3j+k) \)
Answer: We will find the shortest distance between these lines using the vector method. First, we identify the components of each line equation, which are in the form \( \vec { r } = \vec { a } + \lambda \vec { b } \).
For the first line:
\( \vec { a_1 } = i+2j+3k \)
\( \vec { b_1 } = i-3j+2k \)
For the second line:
\( \vec { a_2 } = 4i+5j+6k \)
\( \vec { b_2 } = 2i+3j+k \)
Next, we calculate the vector connecting the two points \( \vec { a_2 } - \vec { a_1 } \):
\( \vec { a_2 } - \vec { a_1 } = (4i+5j+6k) - (i+2j+3k) \)
\( = (4-1)i + (5-2)j + (6-3)k \)
\( = 3i+3j+3k \)
Then, we find the cross product of the direction vectors \( \vec { b_1 } \times \vec { b_2 } \):
\( \vec { b_1 } \times \vec { b_2 } = (i-3j+2k) \times (2i+3j+k) \)
We use a determinant to compute this cross product:
\[ \begin{vmatrix} i & j & k \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} \] \( = i((-3)(1) - (2)(3)) - j((1)(1) - (2)(2)) + k((1)(3) - (-3)(2)) \)
\( = i(-3-6) - j(1-4) + k(3+6) \)
\( = -9i - (-3)j + 9k \)
\( = -9i+3j+9k \)
Now, we find the magnitude of the cross product \( | \vec { b_1 } \times \vec { b_2 } | \):
\( | \vec { b_1 } \times \vec { b_2 } | = \sqrt{(-9)^2 + (3)^2 + (9)^2} \)
\( = \sqrt{81+9+81} \)
\( = \sqrt{171} \)
To simplify \( \sqrt{171} \), we can factor it as \( \sqrt{9 \times 19} = 3\sqrt{19} \).
Finally, we calculate the shortest distance \( d \) using the formula:
\( d = \left| \frac { (\vec { a_2 } - \vec { a_1 }) \cdot (\vec { b_1 } \times \vec { b_2 }) }{ | \vec { b_1 } \times \vec { b_2 } | } \right| \)
First, the dot product \( (\vec { a_2 } - \vec { a_1 }) \cdot (\vec { b_1 } \times \vec { b_2 }) \):
\( (3i+3j+3k) \cdot (-9i+3j+9k) \)
\( = (3)(-9) + (3)(3) + (3)(9) \)
\( = -27 + 9 + 27 \)
\( = 9 \)
Substitute these values into the distance formula:
\( d = \left| \frac { 9 }{ 3\sqrt{19} } \right| \)
\( d = \frac { 3 }{ \sqrt{19} } \)
To rationalize the denominator, multiply by \( \frac { \sqrt{19} }{ \sqrt{19} } \):
\( d = \frac { 3\sqrt{19} }{ 19 } \)
The shortest distance between the lines is \( \frac{3\sqrt{19}}{19} \). This method is widely used for finding distances between skew lines.
In simple words: We take the two line equations and pick out their starting points and direction vectors. We subtract the starting points and then cross-multiply the direction vectors. After that, we perform a dot product and divide by the length of the cross-product to get the smallest distance between the lines.

🎯 Exam Tip: Pay close attention to the vector operations, especially the cross product and dot product, as a small error in calculation can lead to a completely different result.

Question 4. Find the shortest distance between the lines, whose vector equations are
\( \vec { r } =(1-t)i+(t-2)j+(3-2t)k \)
and
\( \vec { r } =(s+1)i+(2s-1)j-(2s+1)k \)
Answer: First, we rewrite the given vector equations in the standard form \( \vec { r } = \vec { a } + \lambda \vec { b } \).
For the first line:
\( \vec { r } = i-ti + tj-2j + 3k-2tk \)
\( \vec { r } = (i-2j+3k) + t(-i+j-2k) \)
So, \( \vec { a_1 } = i-2j+3k \) and \( \vec { b_1 } = -i+j-2k \).
For the second line:
\( \vec { r } = si+i + 2sj-j - 2sk-k \)
\( \vec { r } = (i-j-k) + s(i+2j-2k) \)
So, \( \vec { a_2 } = i-j-k \) and \( \vec { b_2 } = i+2j-2k \).
Next, we calculate the vector connecting the two points \( \vec { a_2 } - \vec { a_1 } \):
\( \vec { a_2 } - \vec { a_1 } = (i-j-k) - (i-2j+3k) \)
\( = (1-1)i + (-1-(-2))j + (-1-3)k \)
\( = 0i + ( -1+2 )j - 4k \)
\( = j-4k \)
Then, we find the cross product of the direction vectors \( \vec { b_1 } \times \vec { b_2 } \):
\( \vec { b_1 } \times \vec { b_2 } = (-i+j-2k) \times (i+2j-2k) \)
We use a determinant for this calculation:
\[ \begin{vmatrix} i & j & k \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} \] \( = i((1)(-2) - (-2)(2)) - j((-1)(-2) - (-2)(1)) + k((-1)(2) - (1)(1)) \)
\( = i(-2+4) - j(2+2) + k(-2-1) \)
\( = 2i - 4j - 3k \)
Now, we find the magnitude of the cross product \( | \vec { b_1 } \times \vec { b_2 } | \):
\( | \vec { b_1 } \times \vec { b_2 } | = \sqrt{(2)^2 + (-4)^2 + (-3)^2} \)
\( = \sqrt{4+16+9} \)
\( = \sqrt{29} \)
Finally, we calculate the shortest distance \( d \) using the formula:
\( d = \left| \frac { (\vec { a_2 } - \vec { a_1 }) \cdot (\vec { b_1 } \times \vec { b_2 }) }{ | \vec { b_1 } \times \vec { b_2 } | } \right| \)
First, calculate the dot product \( (\vec { a_2 } - \vec { a_1 }) \cdot (\vec { b_1 } \times \vec { b_2 }) \):
\( (j-4k) \cdot (2i-4j-3k) \)
\( = (0)(2) + (1)(-4) + (-4)(-3) \)
\( = 0 - 4 + 12 \)
\( = 8 \)
Substitute these values into the distance formula:
\( d = \left| \frac { 8 }{ \sqrt{29} } \right| \)
\( d = \frac { 8 }{ \sqrt{29} } \)
To rationalize the denominator, multiply by \( \frac { \sqrt{29} }{ \sqrt{29} } \):
\( d = \frac { 8\sqrt{29} }{ 29 } \)
The shortest distance between the lines is \( \frac{8\sqrt{29}}{29} \). Rewriting vector equations in standard form is crucial for these types of problems.
In simple words: First, we change the line equations to a standard format that shows a starting point and a direction vector for each. Then we find the vector between the two starting points and the cross-product of their direction vectors. Using a special formula, we take the dot product of these two new vectors and divide it by the length of the cross-product to get the shortest distance.

🎯 Exam Tip: When lines are given in a non-standard parametric form like \( (1-t)i \), always convert them to \( \vec{a} + t\vec{b} \) by separating the constant terms from the 't' terms first. This helps identify \( \vec{a} \) and \( \vec{b} \) correctly.

Question 5. Find the shortest distance between the lines,
\( \frac { x-1 }{ 2 } = \frac { y+1 }{ 3 } = z \)
and
\( \frac { x+1 }{ 3 } = \frac { y-2 }{ 1 } = \frac { z-2 }{ 0 } \)
Answer: We have two lines given in Cartesian form. To find the shortest distance, we'll use a method that involves finding a common perpendicular between the lines. Let the first line be (1) and the second line be (2).
Line (1): \( \frac { x-1 }{ 2 } = \frac { y+1 }{ 3 } = \frac { z-0 }{ 1 } = r_1 \) (let's say)
Line (2): \( \frac { x+1 }{ 3 } = \frac { y-2 }{ 1 } = \frac { z-2 }{ 0 } = r_2 \) (let's say)
Any point on line (1), say P, can be written as \( P(2r_1+1, 3r_1-1, r_1) \).
Any point on line (2), say Q, can be written as \( Q(3r_2-1, r_2+2, r_2+2) \).
The direction ratios (DR's) of the line segment PQ are:
\( (3r_2-1-(2r_1+1)), (r_2+2-(3r_1-1)), (r_2+2-r_1) \)
\( = (3r_2-2r_1-2), (r_2-3r_1+3), (r_2-r_1+2) \)
Since PQ is the shortest distance, it is perpendicular to both line (1) and line (2).
If PQ is perpendicular to line (1), the dot product of their direction ratios is zero:
\( 2(3r_2-2r_1-2) + 3(r_2-3r_1+3) + 1(r_2-r_1+2) = 0 \)
\( 6r_2-4r_1-4 + 3r_2-9r_1+9 + r_2-r_1+2 = 0 \)
\( (6+3+1)r_2 + (-4-9-1)r_1 + (-4+9+2) = 0 \)
\( 10r_2 - 14r_1 + 7 = 0 \) ... (3)
If PQ is perpendicular to line (2), the dot product of their direction ratios is zero:
\( 3(3r_2-2r_1-2) + 1(r_2-3r_1+3) + 0(r_2-r_1+2) = 0 \)
\( 9r_2-6r_1-6 + r_2-3r_1+3 + 0 = 0 \)
\( (9+1)r_2 + (-6-3)r_1 + (-6+3) = 0 \)
\( 10r_2 - 9r_1 - 3 = 0 \) ... (4)
Now we solve equations (3) and (4) simultaneously for \( r_1 \) and \( r_2 \).
From (4), \( 10r_2 = 9r_1+3 \)
Substitute \( 10r_2 \) into (3):
\( (9r_1+3) - 14r_1 + 7 = 0 \)
\( -5r_1 + 10 = 0 \)
\( -5r_1 = -10 \)
\( r_1 = 2 \)
Substitute \( r_1 = 2 \) into \( 10r_2 = 9r_1+3 \):
\( 10r_2 = 9(2)+3 \)
\( 10r_2 = 18+3 \)
\( 10r_2 = 21 \)
\( r_2 = \frac { 21 }{ 10 } \)
Now we find the coordinates of points P and Q using \( r_1=2 \) and \( r_2=\frac{21}{10} \).
For point P on line (1):
\( x_P = 2(2)+1 = 5 \)
\( y_P = 3(2)-1 = 5 \)
\( z_P = 2 \)
So, \( P(5, 5, 2) \).
For point Q on line (2):
\( x_Q = 3\left(\frac{21}{10}\right)-1 = \frac{63}{10}-1 = \frac{53}{10} \)
\( y_Q = \frac{21}{10}+2 = \frac{21+20}{10} = \frac{41}{10} \)
\( z_Q = \frac{21}{10}+2 = \frac{41}{10} \)
So, \( Q\left(\frac{53}{10}, \frac{41}{10}, \frac{41}{10}\right) \).
Finally, we calculate the shortest distance PQ using the distance formula between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \):
\( PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \)
\( PQ = \sqrt{\left(\frac{53}{10}-5\right)^2 + \left(\frac{41}{10}-5\right)^2 + \left(\frac{41}{10}-2\right)^2} \)
\( PQ = \sqrt{\left(\frac{53-50}{10}\right)^2 + \left(\frac{41-50}{10}\right)^2 + \left(\frac{41-20}{10}\right)^2} \)
\( PQ = \sqrt{\left(\frac{3}{10}\right)^2 + \left(\frac{-9}{10}\right)^2 + \left(\frac{21}{10}\right)^2} \)
\( PQ = \sqrt{\frac{9}{100} + \frac{81}{100} + \frac{441}{100}} \)
\( PQ = \sqrt{\frac{9+81+441}{100}} \)
\( PQ = \sqrt{\frac{531}{100}} \)
\( PQ = \frac{\sqrt{531}}{10} \)
We can simplify \( \sqrt{531} = \sqrt{9 \times 59} = 3\sqrt{59} \).
So, \( PQ = \frac{3\sqrt{59}}{10} \). This detailed method allows us to find the exact points on each line that are closest to each other.
In simple words: First, we find a general point on each line using variables. Since the shortest line segment connecting them must be perpendicular to both lines, we use this condition to create two equations. Solving these equations gives us the specific points on each line that are closest. Then, we just calculate the distance between these two points.

🎯 Exam Tip: This method of finding points P and Q and then the distance is generally used for non-parallel skew lines. Remember that if the lines are parallel, a different (simpler) formula for distance is used.