The lines \(\frac{1-x}{2}=\frac{y-1}{3}=\frac{z}{1}\) and \(\frac{2x-3}{2p}=\frac{y}{-1}=\frac{z-4}{7}\) are perpendicular to each other for p equal to:
Rewrite the equations of the lines in standard form:
Line 1: \(\frac{x-1}{-2}=\frac{y-1}{3}=\frac{z}{1}\). Direction ratios: \((-2, 3, 1)\)
Line 2: \(\frac{x-\frac{3}{2}}{p}=\frac{y}{-1}=\frac{z-4}{7}\). Direction ratios: \((p, -1, 7)\)
Apply the condition for perpendicularity:
If two lines with direction ratios \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\) are perpendicular, then \(a_1a_2 + b_1b_2 + c_1c_2 = 0\).
So, \((-2)(p) + (3)(-1) + (1)(7) = 0\)
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Pedagogical Audit
Bloom's Analysis:
This is an APPLY question because the student needs to apply the condition of perpendicularity of two lines in 3D geometry to find the value of 'p'. This involves manipulating the given equations into standard form and then using the formula.
Knowledge Dimension:PROCEDURAL
Justification:The question requires the student to follow a specific procedure: converting the line equations to standard form, identifying direction ratios, applying the perpendicularity condition, and solving for 'p'.
Syllabus Audit:
In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of 3D geometry, specifically the condition for perpendicularity of lines, which is a standard concept covered in the textbook.