Home
Class 12
MATHS
Let A(1, 1, 1), B(3, 7, 4) and C (-1, 3,...

Let A(1, 1, 1), B(3, 7, 4) and C (-1, 3, 0) be three
points in a plane.
The perpendicular distance from C to AB is

A

`(sqrt(25312))/49`

B

`(sqrt(23571))/49`

C

`(14321)^(1/2)/49`

D

`(sqrt(20384))/49`

Text Solution

AI Generated Solution

The correct Answer is:
To find the perpendicular distance from point C to the line segment AB, we can follow these steps: ### Step 1: Determine the direction ratios of line AB Given points A(1, 1, 1) and B(3, 7, 4), we can find the direction ratios of line AB. The direction ratios can be calculated as follows: \[ \text{Direction ratios of } AB = (B_x - A_x, B_y - A_y, B_z - A_z) = (3 - 1, 7 - 1, 4 - 1) = (2, 6, 3) \] ### Step 2: Write the parametric equations of line AB Using point A and the direction ratios, we can write the parametric equations for the line AB: \[ x = 1 + 2\lambda, \quad y = 1 + 6\lambda, \quad z = 1 + 3\lambda \] ### Step 3: Determine a point Q on line AB Let Q be a point on line AB represented by the parameter \(\lambda\): \[ Q(\lambda) = (1 + 2\lambda, 1 + 6\lambda, 1 + 3\lambda) \] ### Step 4: Calculate the direction ratios of line QC The direction ratios of line QC from point C(-1, 3, 0) to point Q are: \[ \text{Direction ratios of } QC = (Q_x - C_x, Q_y - C_y, Q_z - C_z) = ((1 + 2\lambda) - (-1), (1 + 6\lambda) - 3, (1 + 3\lambda) - 0) \] \[ = (2\lambda + 2, 6\lambda - 2, 3\lambda + 1) \] ### Step 5: Set up the condition for perpendicularity For line AB and line QC to be perpendicular, the dot product of their direction ratios must equal zero: \[ (2, 6, 3) \cdot (2\lambda + 2, 6\lambda - 2, 3\lambda + 1) = 0 \] Calculating the dot product: \[ 2(2\lambda + 2) + 6(6\lambda - 2) + 3(3\lambda + 1) = 0 \] \[ 4\lambda + 4 + 36\lambda - 12 + 9\lambda + 3 = 0 \] \[ (4 + 36 + 9)\lambda + (4 - 12 + 3) = 0 \] \[ 49\lambda - 5 = 0 \] \[ 49\lambda = 5 \quad \Rightarrow \quad \lambda = \frac{5}{49} \] ### Step 6: Find the coordinates of point Q Substituting \(\lambda = \frac{5}{49}\) back into the parametric equations for Q: \[ Q = \left(1 + 2\left(\frac{5}{49}\right), 1 + 6\left(\frac{5}{49}\right), 1 + 3\left(\frac{5}{49}\right)\right) \] \[ = \left(1 + \frac{10}{49}, 1 + \frac{30}{49}, 1 + \frac{15}{49}\right) \] \[ = \left(\frac{59}{49}, \frac{79}{49}, \frac{64}{49}\right) \] ### Step 7: Calculate the distance QC Now, we calculate the distance from point C to point Q using the distance formula: \[ d = \sqrt{(Q_x - C_x)^2 + (Q_y - C_y)^2 + (Q_z - C_z)^2} \] \[ = \sqrt{\left(\frac{59}{49} - (-1)\right)^2 + \left(\frac{79}{49} - 3\right)^2 + \left(\frac{64}{49} - 0\right)^2} \] \[ = \sqrt{\left(\frac{59}{49} + \frac{49}{49}\right)^2 + \left(\frac{79}{49} - \frac{147}{49}\right)^2 + \left(\frac{64}{49}\right)^2} \] \[ = \sqrt{\left(\frac{108}{49}\right)^2 + \left(-\frac{68}{49}\right)^2 + \left(\frac{64}{49}\right)^2} \] \[ = \sqrt{\frac{11664}{2401} + \frac{4624}{2401} + \frac{4096}{2401}} \] \[ = \sqrt{\frac{11664 + 4624 + 4096}{2401}} = \sqrt{\frac{20484}{2401}} = \frac{\sqrt{20484}}{49} \] ### Final Answer Thus, the perpendicular distance from point C to line AB is: \[ \frac{\sqrt{20484}}{49} \text{ units} \]
Promotional Banner

Topper's Solved these Questions

  • THREE DIMENSIONAL GEOMETRY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION - D COMPREHENSION IV|3 Videos
  • THREE DIMENSIONAL GEOMETRY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION - D COMPREHENSION V|3 Videos
  • THREE DIMENSIONAL GEOMETRY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION - D COMPREHENSION II|3 Videos
  • STRAIGHT LINES

    AAKASH INSTITUTE ENGLISH|Exercise SECTION-J (AAKASH CHALLENGERS QUESTIONS)|5 Videos
  • TRIGNOMETRIC FUNCTIONS

    AAKASH INSTITUTE ENGLISH|Exercise Section - J (Akash Challengers Question)|16 Videos

Similar Questions

Explore conceptually related problems

Let A(1, 1, 1), B(3, 7, 4) and C (-1, 3, 0) be three points in a plane. The co-ordinates of the foot of perpendicular drawn from point C to the line segment AB is

Let A(1, 1, 1), B(3, 7, 4) and C (-1, 3, 0) be three points in a plane. Area of the triangle ABC is (in sq. units)

Let A(1,1,1),B(2,3,5)a n dC(-1,0,2) be three points, then equation of a plane parallel to the plane A B C which is at distance 2units is

Let A(1, 2, 3), B(0, 0, 1), C(-1, 1, 1) are the vertices of a DeltaABC .The equation of median through C to side AB is

Let A(1,1,1),B(2,3,5)a n dC(-1,0,2) be three points, then equation of a plane parallel to the plane A B C which is at distance 2 is a. 2x-3y+z+2sqrt(14)=0 b. 2x-3y+z-sqrt(14)=0 c. 2x-3y+z+2=0 d. 2x-3y+z-2=0

A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular form AonB Cdot Find the coordinates of Ddot

Let A(6, -1), B (1, 3) and C (x, 8) be three points such that AB = BC then the value of x are

Given four points A(2, 1, 0), B(1, 0, 1), C(3, 0, 1) and D(0, 0, 2) . Point D lies on a line L orthogonal to the plane determined by the points A, B and C. Q. The perpendicular distance of D from the plane ABC is

Let there are three points A (0, 4/3), B(–1, 0) and C(1, 0) in x – y plane. The distance from a variable point P to the line BC is the geometic mean of the distances from this point to lines AB and AC then locus of P can be

Find the equation of the plane determined by the points A(3, –1, 2), B(5, 2, 4) and C(–1, –1, 6) and hence find the distance between the plane and the point P(6, 5, 9).