Question

If A(2,1,-1), B(6,-3,2), C(-3,12,4) are the vertices of a triangle ABC and the equation of the plane containing the triangle ABC is $53x+by+cz+d=0$, then $\frac{d}{b+c}=$

• A. -5
• B. 1
• C. 4
• D. -15

Hint:

The equation of a plane passing through three points A, B, and C can also be found using the scalar triple product in determinant form: $\begin{vmatrix} x-x_A & y-y_A & z-z_A
x_B-x_A & y_B-y_A & z_B-z_A
x_C-x_A & y_C-y_A & z_C-z_A \end{vmatrix} = 0$. This is equivalent to finding the cross product for the normal and then writing the plane equation, and can sometimes be a faster setup.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The equation of a plane containing three non-collinear points A, B, and C can be determined by finding a normal vector to the plane. A normal vector is perpendicular to any two non-parallel vectors lying in the plane, such as \( \vec{AB} \) and \( \vec{AC} \). This normal vector can be found by calculating the cross product of these two vectors. Once the normal vector is found, the equation of the plane can be written using one of the given points.
Step 2: Key Formula or Approach
1. Find two vectors in the plane, for example, \( \vec{AB} \) and \( \vec{AC} \). 2. Calculate the normal vector to the plane, \( \vec{n} \), by taking the cross product: \( \vec{n} = \vec{AB} \times \vec{AC} \). 3. The equation of the plane passing through a point \( A(x_0, y_0, z_0) \) with a normal vector \( \vec{n} = A\hat{i} + B\hat{j} + C\hat{k} \) is given by \( A(x-x_0) + B(y-y_0) + C(z-z_0) = 0 \). 4. Compare the derived equation with the given form \( 53x + by + cz + d = 0 \) to find the values of \( b, c, \) and \( d \). 5. Calculate the final expression \( \frac{d}{b+c} \).
Step 3: Detailed Explanation
The given vertices of the triangle are A(2, 1, -1), B(6, -3, 2), and C(-3, 12, 4). 1. Find two vectors in the plane: \[ \vec{AB} = \vec{OB} - \vec{OA} = (6-2)\hat{i} + (-3-1)\hat{j} + (2-(-1))\hat{k} = 4\hat{i} - 4\hat{j} + 3\hat{k} \] \[ \vec{AC} = \vec{OC} - \vec{OA} = (-3-2)\hat{i} + (12-1)\hat{j} + (4-(-1))\hat{k} = -5\hat{i} + 11\hat{j} + 5\hat{k} \] 2. Find the normal vector \( \vec{n} \): \[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -4 & 3 \\ -5 & 11 & 5 \end{vmatrix} \] \[ \vec{n} = \hat{i}((-4)(5) - (3)(11)) - \hat{j}((4)(5) - (3)(-5)) + \hat{k}((4)(11) - (-4)(-5)) \] \[ \vec{n} = \hat{i}(-20 - 33) - \hat{j}(20 + 15) + \hat{k}(44 - 20) \] \[ \vec{n} = -53\hat{i} - 35\hat{j} + 24\hat{k} \] 3. Find the equation of the plane: The normal vector is \( \vec{n} = (-53, -35, 24) \). We can use point A(2, 1, -1) to write the equation of the plane. \[ -53(x-2) - 35(y-1) + 24(z-(-1)) = 0 \] \[ -53x + 106 - 35y + 35 + 24z + 24 = 0 \] \[ -53x - 35y + 24z + 165 = 0 \] 4. Compare with the given form: The given equation is \( 53x + by + cz + d = 0 \). To match the coefficient of \( x \), we multiply our equation by -1. \[ 53x + 35y - 24z - 165 = 0 \] Comparing this with \( 53x + by + cz + d = 0 \), we get: \( b = 35 \) \( c = -24 \) \( d = -165 \) 5. Calculate the final expression: \[ \frac{d}{b+c} = \frac{-165}{35 + (-24)} = \frac{-165}{11} = -15 \] Step 4: Final Answer
The value of the expression \( \frac{d}{b+c} \) is -15.