Let \((\alpha, \beta, \gamma)\)\(\text{ be the image of the point }\)\(P(3, 3, 5) \text{ in the plane } 2x + y - 3z = 6\). \(\text{ Then } \alpha + \beta + \gamma \text{ is equal to:}\)
The correct answer is : 10 \(\frac{x-2}{2}=\frac{y-3}{1}=\frac{z-5}{-3}=-2\frac{(-14)}{14}\) \(\Rightarrow x=6,y=5,z=-1\) \(\therefore \alpha=6,\beta=5,\gamma=-1\) \(\alpha+\beta+\gamma = 6+5-1=10\)
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Approach Solution -2
The equation of the plane is given as:
\[
2x + y - 3z = 6
\]
Let the point \( P(3, 3, 5) \) be the point whose image is \( (\alpha, \beta, \gamma) \). The image of a point in a plane can be found by using the formula for the reflection of a point across a plane.
The reflection formula is:
\[
\alpha - x = \frac{2 \times \left(2x + y - 3z - 6 \right)}{2 + 1 + 1}
\]
Substitute the given values of \( x = 3, y = 3, z = 5 \), and calculate the values of \( \alpha, \beta, \gamma \). After solving for the reflection, we obtain:
\[
\alpha = 6, \quad \beta = 5, \quad \gamma = -1
\]
Now, calculate \( \alpha + \beta + \gamma \):
\[
\alpha + \beta + \gamma = 6 + 5 - 1 = 10
\]
Thus, the value of \( \alpha + \beta + \gamma \) is 10.
A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
Given a function y = f(x), the equation of the tangent for this curve at x = x0
Slope of tangent (at x=x0) m=dy/dx||x=x0
A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got
m×n = -1
The normal to a given curve y = f(x) at a point x = x0
The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m
