Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1396
COMPETENCY BASED
REMEMBER
5 Marks 2025 AISSCE(Board Exam) LA
Find the image A' of the point A(2, 1, 2) in the line $l:\vec{r}=4\hat{i}+2\hat{j}+2\hat{k}+\lambda(\hat{i}-\hat{j}-\hat{k})$. Also, find the equation of line joining AA'. Find the foot of perpendicular from point A on the line l.
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Detailed Solution<\/h3>\r\n <\/div>\r\n\r\n
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Step 1: Find the coordinates of a general point P on the line l.

The equation of the line $l$ is given by $\vec{r}=4\hat{i}+2\hat{j}+2\hat{k}+\lambda(\hat{i}-\hat{j}-\hat{k})$. A general point P on this line can be represented as $P(4+\lambda, 2-\lambda, 2-\lambda)$.

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Step 2: Express the direction ratios of AP.

The coordinates of point A are (2, 1, 2). The direction ratios of the line segment AP are given by $(4+\lambda - 2, 2-\lambda - 1, 2-\lambda - 2)$, which simplifies to $(\lambda+2, 1-\lambda, -\lambda)$.

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Step 3: Use the perpendicularity condition.

Since AP is perpendicular to the line $l$, the dot product of their direction ratios must be zero. The direction ratios of line $l$ are (1, -1, -1). Therefore,\r\n$(\lambda+2)(1) + (1-\lambda)(-1) + (-\lambda)(-1) = 0$\r\n$\lambda + 2 - 1 + \lambda + \lambda = 0$\r\n$3\lambda + 1 = 0$\r\n$\lambda = -\frac{1}{3}$

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Step 4: Find the coordinates of the foot of the perpendicular (P).

Substitute $\lambda = -\frac{1}{3}$ into the coordinates of point P: $P(4-\frac{1}{3}, 2+\frac{1}{3}, 2+\frac{1}{3}) = P(\frac{11}{3}, \frac{7}{3}, \frac{7}{3})$.

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Step 5: Find the coordinates of the image A'.

Since P is the midpoint of AA', let A' be $(x, y, z)$. Then,\r\n$\frac{x+2}{2} = \frac{11}{3}$, $\frac{y+1}{2} = \frac{7}{3}$, $\frac{z+2}{2} = \frac{7}{3}$\r\n$x+2 = \frac{22}{3}$, $y+1 = \frac{14}{3}$, $z+2 = \frac{14}{3}$\r\n$x = \frac{22}{3} - 2 = \frac{16}{3}$, $y = \frac{14}{3} - 1 = \frac{11}{3}$, $z = \frac{14}{3} - 2 = \frac{8}{3}$\r\nSo, A' is $(\frac{16}{3}, \frac{11}{3}, \frac{8}{3})$.

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Step 6: Find the equation of the line joining AA'.

The direction ratios of the line AA' are $(\frac{16}{3} - 2, \frac{11}{3} - 1, \frac{8}{3} - 2) = (\frac{10}{3}, \frac{8}{3}, \frac{2}{3})$. We can use the direction ratios (5, 4, 1).\r\nThe equation of the line AA' passing through A(2, 1, 2) is given by:\r\n$\frac{x-2}{5} = \frac{y-1}{4} = \frac{z-2}{1}$

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\r\n Final Answer: Image A' = $(\frac{16}{3}, \frac{11}{3}, \frac{8}{3})$, Equation of line AA': $\frac{x-2}{5} = \frac{y-1}{4} = \frac{z-2}{1}$, Foot of perpendicular P = $(\frac{11}{3}, \frac{7}{3}, \frac{7}{3})$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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\r\n Pedagogical Audit<\/span>\r\n <\/div>\r\n
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\r\n Bloom's Analysis:<\/span> \r\n This is an REMEMBER<\/span> question because it requires recalling the formula for finding the image of a point in a line and the equation of a line in 3D space.\r\n <\/div>\r\n
\r\n Knowledge Dimension:<\/span> \r\n PROCEDURAL<\/span>\r\n
\r\n Justification:<\/span> \r\n The question requires a series of steps to be performed in a specific order, such as finding the general point on a line, using the perpendicularity condition, and applying the midpoint formula.<\/span>\r\n <\/div>\r\n <\/div>\r\n
\r\n Syllabus Audit:<\/span> \r\n In the context of CBSE Class 12, this is classified as COMPETENCY<\/span>. It requires the application of concepts related to 3D geometry, including lines and points in space, and the ability to solve problems involving perpendicularity and images.<\/span>. It assesses the student's ability to apply these concepts in a problem-solving context.\r\n <\/div>\r\n <\/div>\r\n <\/div>

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