CBSE Class 12 Sample Paper for 2026 Boards

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Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Question 14 - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at August 12, 2026 by

The distance of the point with position vector 3ı ˆ+4ȷ ˆ+5𝑘 ˆ from - CBSE Class 12 Sample Paper for 2026 Boards

part 2 - Question 14 - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 14 The distance of the point with position vector 3𝚤 ˆ+4𝚥 ˆ+5𝑘 ˆ from the y -axis is (A) 4 units (B) √34 units (C) 5 units (D) 5√2 unitsFormula To find the distance of a point P with coordinates (𝒙,𝒚,𝒛) from one of the coordinate axes, we use the following formulas for the perpendicular distance: Distance from the 𝒙-axis =√(𝑦^2+𝑧^2 ) Distance from the 𝒚-axis =√(𝑥^2+𝑧^2 ) Distance from the 𝒛-axis =√(𝑥^2+𝑦^2 ) For this question, we need formula for distance from the 𝐲-axis. point is on y-axis its x & z coordinate is 0 Let the point on y-axis be A(0, a, 0) Given that We need to find stance of Point A is at a distance of 5√2 from point P (3, – 2, 5) i.e. PA = 5√2 Finding Distance PA Since the a point is on the y-axis, then the coordinate of x and y are 0. So, the point is (0, y, 0) The distance of the point with position vector 3ı ˆ+4ȷ ˆ+5𝑘 ˆ from the y -axis is So, the correct answer is (B)

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