Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1332
COMPETENCY BASED
REMEMBER
5 Marks 2024 AISSCE(Board Exam) LA
If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-7}$ are perpendicular to each other, find the value of k and hence write the vector equation of a line perpendicular to these two lines and passing through the point (3, -4, 7).
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Detailed Solution<\/h3>\r\n <\/div>\r\n\r\n
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Step 1: Find the direction ratios of the two lines

The direction ratios of the first line are -3, 2k, 2 and the direction ratios of the second line are 3k, 1, -7.

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Step 2: Apply the perpendicularity condition

Since the lines are perpendicular, the dot product of their direction ratios is zero: (-3)(3k) + (2k)(1) + (2)(-7) = 0

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Step 3: Solve for k

Simplifying the equation: -9k + 2k - 14 = 0 => -7k = 14 => k = -2

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Step 4: Find the direction ratios of the perpendicular line

The direction ratios of the first line are -3, -4, 2 (since k = -2) and the direction ratios of the second line are -6, 1, -7 (since k = -2). Let the direction ratios of the line perpendicular to both lines be a, b, c. Then, -3a - 4b + 2c = 0 and -6a + b - 7c = 0 We can find the direction ratios by taking the cross product of the direction vectors of the given lines: $$\\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\ -3 & -4 & 2 \\\\ -6 & 1 & -7 \\end{vmatrix} = (28 - 2)\\hat{i} - (21 + 12)\\hat{j} + (-3 - 24)\\hat{k} = 26\\hat{i} - 33\\hat{j} - 27\\hat{k}$$ So, the direction ratios are 26, -33, -27.

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Step 5: Write the vector equation of the line

The line passes through the point (3, -4, 7). The vector equation of the line is: $\\vec{r} = (3\\hat{i} - 4\\hat{j} + 7\\hat{k}) + \\lambda(26\\hat{i} - 33\\hat{j} - 27\\hat{k})$

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\r\n Final Answer: k = -2, $\\vec{r} = (3\\hat{i} - 4\\hat{j} + 7\\hat{k}) + \\lambda(26\\hat{i} - 33\\hat{j} - 27\\hat{k})$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

AI generated content. Review strictly for academic accuracy.

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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 condition for perpendicularity of lines in 3D space and applying the cross product formula.\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 applying a specific procedure (finding direction ratios, applying the perpendicularity condition, solving for k, finding the cross product, and forming the vector equation).<\/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 assesses the student's ability to apply the concepts of 3D geometry and vector algebra to solve a problem involving perpendicular lines and vector equations.\r\n <\/div>\r\n <\/div>\r\n <\/div>

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