RBSE Solutions Class 12 Maths Chapter 14 Three Dimensional Geometry Exercise 14.4

Question 1. Show that the lines \( \frac { x-1 }{ 2 } = \frac { y-2 }{ 3 } = \frac { z-3}{4} \) and \( \frac {x-4}{5} = \frac {y-1 }{2} = z \), intersect. Find their point of intersection.
Answer:
Let the first line be \( L_1 \): \( \frac { x-1 }{ 2 } = \frac { y-2 }{ 3 } = \frac { z-3}{4} = r_1 \)
Any point on \( L_1 \) can be written as \( (2r_1 + 1, 3r_1 + 2, 4r_1 + 3) \).
Let the second line be \( L_2 \): \( \frac {x-4}{5} = \frac {y-1 }{2} = \frac{z}{1} = r_2 \)
Any point on \( L_2 \) can be written as \( (5r_2 + 4, 2r_2 + 1, r_2) \).
For the lines to intersect, these points must be the same for some \( r_1 \) and \( r_2 \).
So, we equate the coordinates:
\( 2r_1 + 1 = 5r_2 + 4 \)
\( \implies 2r_1 - 5r_2 = 3 \) ...(1)
\( 3r_1 + 2 = 2r_2 + 1 \)
\( \implies 3r_1 - 2r_2 = -1 \) ...(2)
\( 4r_1 + 3 = r_2 \)
\( \implies 4r_1 - r_2 = -3 \) ...(3)

Now, we solve equations (1) and (2).
Multiply equation (1) by 2 and equation (2) by 5:
\( 2 \times (2r_1 - 5r_2 = 3) \implies 4r_1 - 10r_2 = 6 \)
\( 5 \times (3r_1 - 2r_2 = -1) \implies 15r_1 - 10r_2 = -5 \)
Subtract the first new equation from the second:
\( (15r_1 - 10r_2) - (4r_1 - 10r_2) = -5 - 6 \)
\( \implies 11r_1 = -11 \)
\( \implies r_1 = -1 \)

Substitute \( r_1 = -1 \) into equation (1):
\( 2(-1) - 5r_2 = 3 \)
\( \implies -2 - 5r_2 = 3 \)
\( \implies -5r_2 = 5 \)
\( \implies r_2 = -1 \)

We must check if these values of \( r_1 \) and \( r_2 \) satisfy the third equation (3):
\( 4r_1 - r_2 = -3 \)
\( 4(-1) - (-1) = -4 + 1 = -3 \)
Since \( -3 = -3 \), the values satisfy all three equations. This means the lines intersect. When \( r_1 \) and \( r_2 \) satisfy all equations, the lines will always cross each other.

Now, we find the point of intersection by substituting \( r_1 = -1 \) into the general point of \( L_1 \), or \( r_2 = -1 \) into the general point of \( L_2 \). Using \( r_1 = -1 \):
\( x = 2(-1) + 1 = -2 + 1 = -1 \)
\( y = 3(-1) + 2 = -3 + 2 = -1 \)
\( z = 4(-1) + 3 = -4 + 3 = -1 \)
The point of intersection is \( (-1, -1, -1) \).
In simple words: To find where two lines meet, we set up equations by making their coordinates equal. If we can find values for the parameters that satisfy all these equations, the lines intersect at that specific point.

🎯 Exam Tip: Always verify the calculated values of \( r_1 \) and \( r_2 \) in all three coordinate equations to confirm that the lines actually intersect. If they don't satisfy the third equation, the lines are skew, not intersecting.

Question 2. Check, whether following lines intersect or not \( \vec { r } = (\hat { i }-\hat { j })+\lambda (2\hat { i }+\hat { k }) \) and \( \vec { r } = (2\hat { i }-\hat { j })+\mu (\hat { i }+\hat { j }-\hat { k }) \)
Answer:
For the lines to intersect, there must be values of \( \lambda \) and \( \mu \) such that the position vectors are equal:
\( (\hat { i }-\hat { j })+\lambda (2\hat { i }+\hat { k }) = (2\hat { i }-\hat { j })+\mu (\hat { i }+\hat { j }-\hat { k }) \)
Expand both sides:
\( (1+2\lambda)\hat{i} + (-1)\hat{j} + (\lambda)\hat{k} = (2+\mu)\hat{i} + (-1+\mu)\hat{j} + (-\mu)\hat{k} \)
Comparing the coefficients of \( \hat{i}, \hat{j}, \hat{k} \):
For \( \hat{i} \): \( 1 + 2\lambda = 2 + \mu \)
\( \implies 2\lambda - \mu = 1 \) ...(1)
For \( \hat{j} \): \( -1 = -1 + \mu \)
\( \implies \mu = 0 \) ...(2)
For \( \hat{k} \): \( \lambda = -\mu \) ...(3)

From equation (2), we immediately get \( \mu = 0 \).
Now, substitute \( \mu = 0 \) into equation (3):
\( \lambda = -(0) \)
\( \implies \lambda = 0 \)

Finally, we must check if these values, \( \lambda = 0 \) and \( \mu = 0 \), satisfy the first equation (1):
\( 2\lambda - \mu = 1 \)
\( 2(0) - (0) = 0 - 0 = 0 \)
Since \( 0 \neq 1 \), the values of \( \lambda \) and \( \mu \) do not satisfy all three equations. This means that no common point exists for these lines. If the parameter values don't work for all parts, the lines cannot meet.
Therefore, the lines do not intersect each other.
In simple words: We try to find if there's a specific time (parameter value) when both lines are at the same spot. If the "times" we find don't match up for all directions, it means the lines never cross.

🎯 Exam Tip: When checking for line intersection in vector form, always solve for the parameters using two equations and then test those values in the third equation. If the third equation is not satisfied, the lines do not intersect.

Question 3. Find the foot of perpendicular drawn from point (2, 3, 4) to the line \( \frac {4-x }{ 2 } = \frac {y}{6} =\frac {1-z}{3} \). Also find the perpendicular distance from point to line.
Answer:
First, we write the given line in standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
The line is \( \frac {-(x-4) }{ 2 } = \frac {y-0}{6} =\frac {-(z-1)}{3} \)
\( \implies \frac {x-4 }{ -2 } = \frac {y-0}{6} =\frac {z-1}{-3} = \lambda \)
Let P be the given point (2, 3, 4).
Any general point Q on the line can be written in terms of \( \lambda \) as \( (-2\lambda + 4, 6\lambda, -3\lambda + 1) \).
The direction ratios of the line segment PQ, connecting P(2, 3, 4) to Q, are:
\( x_2 - x_1 = (-2\lambda + 4) - 2 = -2\lambda + 2 \)
\( y_2 - y_1 = (6\lambda) - 3 = 6\lambda - 3 \)
\( z_2 - z_1 = (-3\lambda + 1) - 4 = -3\lambda - 3 \)

The direction ratios of the given line are (-2, 6, -3).
Since PQ is perpendicular to the given line, the dot product of their direction ratios must be zero:
\( (-2\lambda + 2)(-2) + (6\lambda - 3)(6) + (-3\lambda - 3)(-3) = 0 \)
\( 4\lambda - 4 + 36\lambda - 18 + 9\lambda + 9 = 0 \)
Combine the terms with \( \lambda \): \( (4 + 36 + 9)\lambda = 49\lambda \)
Combine the constant terms: \( -4 - 18 + 9 = -13 \)
So, \( 49\lambda - 13 = 0 \)
\( \implies 49\lambda = 13 \)
\( \implies \lambda = \frac{13}{49} \)

Now, we find the coordinates of the foot of the perpendicular Q by substituting \( \lambda = \frac{13}{49} \) into the general point Q:
\( x = -2\left(\frac{13}{49}\right) + 4 = -\frac{26}{49} + \frac{196}{49} = \frac{170}{49} \)
\( y = 6\left(\frac{13}{49}\right) = \frac{78}{49} \)
\( z = -3\left(\frac{13}{49}\right) + 1 = -\frac{39}{49} + \frac{49}{49} = \frac{10}{49} \)
The foot of the perpendicular, Q, is \( \left(\frac{170}{49}, \frac{78}{49}, \frac{10}{49}\right) \). This method helps us find the exact point on the line that is closest to the given point.

Next, we calculate the perpendicular distance PQ using the distance formula between P(2, 3, 4) and Q\( \left(\frac{170}{49}, \frac{78}{49}, \frac{10}{49}\right) \):
\( PQ = \sqrt{\left(\frac{170}{49} - 2\right)^2 + \left(\frac{78}{49} - 3\right)^2 + \left(\frac{10}{49} - 4\right)^2} \)
\( PQ = \sqrt{\left(\frac{170 - 98}{49}\right)^2 + \left(\frac{78 - 147}{49}\right)^2 + \left(\frac{10 - 196}{49}\right)^2} \)
\( PQ = \sqrt{\left(\frac{72}{49}\right)^2 + \left(\frac{-69}{49}\right)^2 + \left(\frac{-186}{49}\right)^2} \)
\( PQ = \frac{1}{49}\sqrt{72^2 + (-69)^2 + (-186)^2} \)
\( PQ = \frac{1}{49}\sqrt{5184 + 4761 + 34596} \)
\( PQ = \frac{1}{49}\sqrt{44541} \)
\( PQ = \frac{1}{49} \times 211.047 \approx 4.307 \)
The solution in the OCR seems to use different numbers for the distance calculation: \( \frac{3}{7}\sqrt{101} \). Let's re-evaluate the coordinates and distance in the OCR to match. The OCR calculation for PQ is: \( PQ = \sqrt{\left(\frac{170}{49} - 2\right)^2 + \left(\frac{78}{49} - 3\right)^2 + \left(\frac{10}{49} - 4\right)^2} \) \( PQ = \sqrt{\left(\frac{170 - 98}{49}\right)^2 + \left(\frac{78 - 147}{49}\right)^2 + \left(\frac{10 - 196}{49}\right)^2} \) \( PQ = \sqrt{\left(\frac{72}{49}\right)^2 + \left(\frac{-69}{49}\right)^2 + \left(\frac{-186}{49}\right)^2} \) The OCR solution shows the calculation \( PQ = \frac{3}{7}\sqrt{101} \). Let's see if this matches. \( \frac{3}{7}\sqrt{101} \approx \frac{3}{7} \times 10.05 \approx 4.307 \). This matches my numerical result. The exact form \( \frac{3}{7}\sqrt{101} \) is derived from the steps in the OCR for \( \lambda \). Let's recheck the OCR steps for \( \lambda \): \( (-2\lambda + 2)(-2) + (6\lambda - 3)(6) + (-3\lambda - 3)(-3) = 0 \) \( 4\lambda - 4 + 36\lambda - 18 + 9\lambda + 9 = 0 \) \( 49\lambda - 13 = 0 \implies \lambda = \frac{13}{49} \) The OCR shows: \( 49\lambda = 13 \implies \lambda = \frac{13}{49} \) The calculation for Q coordinates using \( \lambda = \frac{13}{49} \) is correct in OCR. The distance calculation in OCR is: \( PQ = \sqrt{\left(\frac{170}{49}-2\right)^2 + \left(\frac{78}{49}-3\right)^2 + \left(\frac{10}{49}-4\right)^2} \) \( PQ = \sqrt{\left(\frac{72}{49}\right)^2 + \left(\frac{-69}{49}\right)^2 + \left(\frac{-186}{49}\right)^2} \) \( PQ = \frac{1}{49}\sqrt{5184 + 4761 + 34596} = \frac{1}{49}\sqrt{44541} \) It appears the OCR steps for distance calculation do not directly lead to \( \frac{3}{7}\sqrt{101} \). However, if we assume the OCR's final answer for distance is correct, we should use it. The OCR states: \( \frac{3}{7}\sqrt{101} \). Let's confirm if \( \frac{1}{49}\sqrt{44541} \) simplifies to this. \( 44541 = 9 \times 4949 = 9 \times 101 \times 49 \). This is crucial. \( \sqrt{44541} = \sqrt{9 \times 49 \times 101} = \sqrt{9} \times \sqrt{49} \times \sqrt{101} = 3 \times 7 \times \sqrt{101} = 21\sqrt{101} \). So, \( PQ = \frac{1}{49} \times 21\sqrt{101} = \frac{21}{49}\sqrt{101} = \frac{3}{7}\sqrt{101} \). The OCR calculation is correct! I just needed to simplify the square root. My initial calculation of 211.047 was approximate. So, continuing with the distance: \( PQ = \frac{1}{49}\sqrt{72^2 + (-69)^2 + (-186)^2} \)
\( PQ = \frac{1}{49}\sqrt{5184 + 4761 + 34596} = \frac{1}{49}\sqrt{44541} \)
Since \( 44541 = 9 \times 49 \times 101 \), we can simplify the square root:
\( \sqrt{44541} = \sqrt{9 \times 49 \times 101} = 3 \times 7 \times \sqrt{101} = 21\sqrt{101} \)
So, \( PQ = \frac{1}{49} \times 21\sqrt{101} = \frac{21}{49}\sqrt{101} = \frac{3}{7}\sqrt{101} \)
The perpendicular distance from the point to the line is \( \frac{3}{7}\sqrt{101} \) units.
In simple words: First, we find a general spot on the line. Then, we make sure the line connecting our starting point to that general spot is perfectly straight (perpendicular) to the main line. Once we find that special spot, we measure the distance between the two points.

🎯 Exam Tip: Always convert the line equation to its standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \) correctly before finding a general point on it. Misinterpreting signs in the initial equation is a common error.

Question 4. Find the shortest distance between the parallel lines \( \vec { r } = (2\hat { i }+3\hat { j }+2\hat { k })+\lambda (2\hat { i }-3\hat { j }+6\hat { k }) \) and \( \vec { r } = (-2\hat { i }+3\hat { j })+\mu (2\hat { i }-3\hat { j }+6\hat { k }) \).
Answer:
Let the first line be \( L_1 \), with equation \( \vec { r } = \vec { a_1 } + \lambda \vec { b } \).
Here, \( \vec { a_1 } = 2\hat { i }+3\hat { j }+2\hat { k } \) (position vector of a point on \( L_1 \)).
And \( \vec { b } = 2\hat { i }-3\hat { j }+6\hat { k } \) (direction vector of \( L_1 \)).

Let the second line be \( L_2 \), with equation \( \vec { r } = \vec { a_2 } + \mu \vec { b } \).
Here, \( \vec { a_2 } = -2\hat { i }+3\hat { j }+0\hat { k } \) (position vector of a point on \( L_2 \)).
The direction vector for \( L_2 \) is also \( \vec { b } = 2\hat { i }-3\hat { j }+6\hat { k } \). Since both lines have the same direction vector \( \vec{b} \), they are parallel.

The formula for the shortest distance (d) between two parallel lines is:
\( d = \frac{| \vec{b} \times (\vec{a_2} - \vec{a_1}) |}{| \vec{b} |} \)

First, calculate \( (\vec{a_2} - \vec{a_1}) \):
\( \vec{a_2} - \vec{a_1} = (-2\hat{i} + 3\hat{j}) - (2\hat{i} + 3\hat{j} + 2\hat{k}) \)
\( = (-2-2)\hat{i} + (3-3)\hat{j} + (0-2)\hat{k} \)
\( = -4\hat{i} + 0\hat{j} - 2\hat{k} \)

Next, calculate the cross product \( \vec{b} \times (\vec{a_2} - \vec{a_1}) \):
\( \vec{b} \times (\vec{a_2} - \vec{a_1}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 6 \\ -4 & 0 & -2 \end{vmatrix} \)
\( = \hat{i}((-3)(-2) - (6)(0)) - \hat{j}((2)(-2) - (6)(-4)) + \hat{k}((2)(0) - (-3)(-4)) \)
\( = \hat{i}(6 - 0) - \hat{j}(-4 - (-24)) + \hat{k}(0 - 12) \)
\( = 6\hat{i} - \hat{j}(-4 + 24) - 12\hat{k} \)
\( = 6\hat{i} - 20\hat{j} - 12\hat{k} \)

Now, find the magnitude of \( | \vec{b} \times (\vec{a_2} - \vec{a_1}) | \):
\( | \vec{b} \times (\vec{a_2} - \vec{a_1}) | = \sqrt{6^2 + (-20)^2 + (-12)^2} \)
\( = \sqrt{36 + 400 + 144} \)
\( = \sqrt{580} \)

Finally, find the magnitude of \( | \vec{b} | \):
\( | \vec{b} | = \sqrt{2^2 + (-3)^2 + 6^2} \)
\( = \sqrt{4 + 9 + 36} \)
\( = \sqrt{49} \)
\( = 7 \)

Substitute these values into the distance formula:
\( d = \frac{\sqrt{580}}{7} \)
\( \implies d = \frac{\sqrt{4 \times 145}}{7} = \frac{2\sqrt{145}}{7} \)
The shortest distance between the two parallel lines is \( \frac{2\sqrt{145}}{7} \) units. This formula helps to quickly find the minimum distance when lines do not cross but run side by side.
In simple words: When two lines are parallel, we pick a point on each line. Then, we find a special vector that shows the direction of the lines. We use a formula that combines these to calculate the smallest gap between the two lines.

🎯 Exam Tip: Remember to use the specific formula for parallel lines (involving \( \vec{b} \times (\vec{a_2} - \vec{a_1}) \)) and not the one for skew lines. Ensure that \( \vec{b_1} = \vec{b_2} \) (or are proportional) to confirm parallelism first.