If the lines (x - 5)/7 = (y + 2)/-5 = (z - 1)/1 and (x - 1)/2 = (y/p)/1 = (z - 3)/-1 are perpendicular, then the value of p will be? Find the area enclosed by the curve y = √x and... If the lines (x - 5)/7 = (y + 2)/-5 = (z - 1)/1 and (x - 1)/2 = (y/p)/1 = (z - 3)/-1 are perpendicular, then the value of p will be? Find the area enclosed by the curve y = √x and y = x.
Understand the Problem
The question consists of two parts: first, determining the value of p such that two given lines are perpendicular; second, finding the area enclosed between the curves y = √x and y = x. The problem requires knowledge of geometry and calculus principles.
Answer
The value of $p$ is $3$.
Answer for screen readers
The value of $p$ is $3$.
Steps to Solve
- Find the direction ratios of the lines
The given lines are represented as:
- Line 1: $$ \frac{x - 5}{7} = \frac{y + 2}{-5} = \frac{z - 1}{1} $$
- Line 2: $$ \frac{x - 1}{2} = \frac{y}{p} = \frac{z - 3}{1} $$
For Line 1, the direction ratios are $(7, -5, 1)$.
For Line 2, the direction ratios are $(2, p, 1)$.
- Use the condition for perpendicular lines
For two lines to be perpendicular, the dot product of their direction ratios must equal zero:
$$ 7 \cdot 2 + (-5) \cdot p + 1 \cdot 1 = 0 $$
- Set up the equation
The equation simplifies to:
$$ 14 - 5p + 1 = 0 $$
- Solve for p
Rearranging gives:
$$ -5p + 15 = 0 $$
Now, solve for $p$:
$$ p = 3 $$
- Find area between curves
The curves given are:
- $$ y = \sqrt{x} $$
- $$ y = x $$
Set the equations equal to find points of intersection:
$$ \sqrt{x} = x $$
- Square both sides
This leads to:
$$ x = x^2 $$
Rearranging gives:
$$ x^2 - x = 0 $$
Factoring results in:
$$ x(x - 1) = 0 $$
Thus, the points of intersection are $x = 0$ and $x = 1$.
- Set up the integral for area
The area between the curves from $0$ to $1$ is given by:
$$ A = \int_{0}^{1} (\sqrt{x} - x) , dx $$
- Calculate the integral
Calculating the integral:
$$ A = \int_{0}^{1} \sqrt{x} , dx - \int_{0}^{1} x , dx $$
The integrals are:
$$ \int \sqrt{x} , dx = \frac{2}{3} x^{3/2} $$
and
$$ \int x , dx = \frac{1}{2} x^2 $$
- Evaluate the definite integrals
Evaluating from $0$ to $1$ gives:
$$ \left[ \frac{2}{3}(1)^{3/2} - \frac{1}{2}(1)^2 \right] - \left[ 0 - 0 \right] $$
Which results in:
$$ \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} $$
The value of $p$ is $3$.
More Information
For two lines to be perpendicular, their direction ratios must satisfy the dot product condition. The area between the curves $y = \sqrt{x}$ and $y = x$ can be found using integration, representing the space between the two functions over the interval of intersection.
Tips
- Failing to set the equations equal to find points of intersection.
- Not properly evaluating integrals or forgetting to apply limits.
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