Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1374
COMPETENCY BASED
UNDERSTAND
5 Marks 2025 AISSCE(Board Exam) LA
Find the image A' of the point $A(1,6,3)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$. Also, find the equation of the line joining A and A'.
Prev Next

AI Tutor Explanation

Powered by Gemini

Detailed Solution

Step 1: Parametric Representation of the Line

Let the given line be $L: \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda$. Any point on this line can be represented as $P(\lambda, 2\lambda+1, 3\lambda+2)$.

Step 2: Midpoint of AA'

Let $A' (x', y', z')$ be the image of $A(1,6,3)$ in the line $L$. Then the midpoint of $AA'$, say $M$, must lie on the line $L$. The coordinates of $M$ are given by $M(\frac{x'+1}{2}, \frac{y'+6}{2}, \frac{z'+3}{2})$.

Step 3: M lies on the Line

Since $M$ lies on the line $L$, we can equate the coordinates of $M$ with the parametric representation of the line: $$ \frac{x'+1}{2} = \lambda, \quad \frac{y'+6}{2} = 2\lambda+1, \quad \frac{z'+3}{2} = 3\lambda+2 $$ This gives us: $$ x' = 2\lambda - 1, \quad y' = 4\lambda - 4, \quad z' = 6\lambda + 1 $$ So, $A' (2\lambda - 1, 4\lambda - 4, 6\lambda + 1)$.

Step 4: Direction Ratios and Perpendicularity

The direction ratios of the line $AA'$ are given by $(2\lambda - 1 - 1, 4\lambda - 4 - 6, 6\lambda + 1 - 3) = (2\lambda - 2, 4\lambda - 10, 6\lambda - 2)$. The direction ratios of the given line $L$ are $(1, 2, 3)$. Since $AA'$ is perpendicular to $L$, the dot product of their direction ratios must be zero: $$ 1(2\lambda - 2) + 2(4\lambda - 10) + 3(6\lambda - 2) = 0 $$ $$ 2\lambda - 2 + 8\lambda - 20 + 18\lambda - 6 = 0 $$ $$ 28\lambda - 28 = 0 $$ $$ \lambda = 1 $$

Step 5: Coordinates of A'

Substituting $\lambda = 1$ into the coordinates of $A'$, we get: $$ A' (2(1) - 1, 4(1) - 4, 6(1) + 1) = A'(1, 0, 7) $$

Step 6: Equation of the Line AA'

The equation of the line joining $A(1,6,3)$ and $A'(1,0,7)$ is given by: $$ \frac{x - 1}{1 - 1} = \frac{y - 6}{0 - 6} = \frac{z - 3}{7 - 3} $$ $$ \frac{x - 1}{0} = \frac{y - 6}{-6} = \frac{z - 3}{4} $$ This implies $x = 1$, and $\frac{y - 6}{-6} = \frac{z - 3}{4}$. Thus, $x = 1$ and $4(y - 6) = -6(z - 3)$, which simplifies to $4y - 24 = -6z + 18$, or $4y + 6z = 42$, or $2y + 3z = 21$.

Final Answer: A'(1, 0, 7) and the equation of the line AA' is x=1, 2y+3z=21

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concepts of 3D geometry, including finding the image of a point in a line and the equation of a line in space. This requires using formulas and procedures in a new context.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a series of steps to solve, including finding the parametric form of a line, finding the midpoint, using the perpendicularity condition, and finally determining the equation of the line. These are all procedural steps.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the concepts of 3D geometry to solve a problem involving finding the image of a point and the equation of a line.

More from this Chapter

VSA
25. (a) Find the vector equation of the line passing through the point $(2, 1, 3)$ and perpendicular to both the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} ; \frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$
MCQ_SINGLE
If P is a point on the line segment joining (3, 6, -1) and (6, 2, -2) and y-coordinate of P is 4, then its z-coordinate is:
LA
Find the shortest distance between the lines $L_{1}$ & $L_{2}$ given below : $L_{1}$: The line passing through (2, -1, 1) and parallel to $\frac{x}{1}=\frac{y}{1}=\frac{z}{3}$ and $L_{2}:\vec{r}=\hat{i}+(2\mu+1)\hat{j}-(\mu+2)\hat{k}$
VSA
25. (b) OR: The equations of a line are $5x-3=15y+7=3-10z$. Write the direction cosines of the line and find the coordinates of a point through which it passes.
LA
Find the equation of the line passing through the point of intersection of the lines $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ and $\frac{x-1}{0}=\frac{y}{-3}=\frac{z-7}{2}$ and perpendicular to these given lines.
View All Questions