A comprehensive platform for Teachers to create standard question papers and Students to practice Case-Based, Assertion-Reason, and Critical Thinking questions.
Create professional PDF/Word papers with logo, instructions, and mixed question types in minutes.
Explore our repository by Class and Topic. Filter by "Knowledge" or "Competency" levels.
For Students. Take timed MCQ tests to check your understanding. Get instant feedback.
According to NEP 2020, rote learning is out. The focus has shifted to assessing a student's ability to apply concepts in real-life situations.
Questions derived from real-world passages to test analytical skills.
Testing the logic behind concepts, not just the definition.
Open-ended scenarios that require thinking beyond the textbook.
We provide complete AI-Powered Explanations for every question.
Let the given line be represented in parametric form. We introduce a parameter $\lambda$ such that:
$\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} = \lambda$
From this, we can express the coordinates of a general point P on the line as:
$x = \lambda - 5$
$y = 4\lambda - 3$
$z = -9\lambda + 6$
So, the coordinates of point P are $(\lambda - 5, 4\lambda - 3, -9\lambda + 6)$.
We are given that the distance between point P and point Q(2, 4, -1) is 7 units. We use the distance formula:
$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$
Substituting the coordinates of P and Q, we get:
$7 = \sqrt{(\lambda - 5 - 2)^2 + (4\lambda - 3 - 4)^2 + (-9\lambda + 6 - (-1))^2}$
$7 = \sqrt{(\lambda - 7)^2 + (4\lambda - 7)^2 + (-9\lambda + 7)^2}$
Squaring both sides of the equation, we get:
$49 = (\lambda - 7)^2 + (4\lambda - 7)^2 + (-9\lambda + 7)^2$
$49 = (\lambda^2 - 14\lambda + 49) + (16\lambda^2 - 56\lambda + 49) + (81\lambda^2 - 126\lambda + 49)$
$49 = 98\lambda^2 - 196\lambda + 147$
$0 = 98\lambda^2 - 196\lambda + 98$
Dividing by 98, we get:
$0 = \lambda^2 - 2\lambda + 1$
The quadratic equation is:
$\lambda^2 - 2\lambda + 1 = 0$
$(\lambda - 1)^2 = 0$
Therefore, $\lambda = 1$.
Substitute $\lambda = 1$ into the coordinates of P:
$x = 1 - 5 = -4$
$y = 4(1) - 3 = 1$
$z = -9(1) + 6 = -3$
So, the coordinates of point P are $(-4, 1, -3)$.
The equation of the line joining two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is given by:
$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$
Here, $P(-4, 1, -3)$ and $Q(2, 4, -1)$. Substituting these values, we get:
$\frac{x - (-4)}{2 - (-4)} = \frac{y - 1}{4 - 1} = \frac{z - (-3)}{-1 - (-3)}$
$\frac{x + 4}{6} = \frac{y - 1}{3} = \frac{z + 3}{2}$
\r\n Final Answer: P(-4, 1, -3) and Equation of line PQ: $\frac{x + 4}{6} = \frac{y - 1}{3} = \frac{z + 3}{2}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>