\(=\) \(\begin{vmatrix} x - 2 & -3 & 4 \\ y + 3 & 4 & -5 \\ z - 1 & -3 & 4 \end{vmatrix} = 0\)
\(x−z−1=0 \)
Distance of \(P(7,−3,−4)\) from Plane is
\(d = \left| \frac{7+4-1}{\sqrt{2}} \right| = 5\sqrt{2}\)
Approach 2:
To find the equation of the plane passing through the three points: Equation of plane: \(\begin{vmatrix} x - 2 & y + 3 & z - 1 -3 & 4 & -3 4 & -5 & 4 \end{vmatrix}\) = 0.
Expanding the determinant: (x - 2)\(\begin{vmatrix} 4 & -3 -5 & 4 \end{vmatrix} - (y + 3)\begin{vmatrix} -3 & -3 4 & 4 \end{vmatrix}\) + (z - 1)\(\begin{vmatrix} -3 & 4 4 & -5 \end{vmatrix}\) = 0.
\[ (x - 2)(16 - 15) - (y + 3)(-12 + 12) + (z - 1)(15 + 12) = 0, \] \[ x - z - 1 = 0. \] The equation of the plane is: \[ x - z - 1 = 0. \] To find the distance of the point \( (7, -3, -4) \) from the plane: \[ \text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}, \] where \( A = 1, B = 0, C = -1, D = -1 \). Substitute \( (x_1, y_1, z_1) = (7, -3, -4) \): \[ \text{Distance} = \frac{|7 + 0 - (-4) - 1|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{|7 + 4 - 1|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}. \]
Show that the following lines intersect. Also, find their point of intersection:
Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]
Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
The vector equations of two lines are given as:
Line 1: \[ \vec{r}_1 = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(4\hat{i} + 6\hat{j} + 12\hat{k}) \]
Line 2: \[ \vec{r}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(6\hat{i} + 9\hat{j} + 18\hat{k}) \]
Determine whether the lines are parallel, intersecting, skew, or coincident. If they are not coincident, find the shortest distance between them.
Determine the vector equation of the line that passes through the point \( (1, 2, -3) \) and is perpendicular to both of the following lines:
\[ \frac{x - 8}{3} = \frac{y + 16}{7} = \frac{z - 10}{-16} \quad \text{and} \quad \frac{x - 15}{3} = \frac{y - 29}{-8} = \frac{z - 5}{-5} \]
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
Mathematically, Geometry is one of the most important topics. The concepts of Geometry are derived w.r.t. the planes. So, Geometry is divided into three major categories based on its dimensions which are one-dimensional geometry, two-dimensional geometry, and three-dimensional geometry.
Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.