Class CBSE Class 12 Mathematics Vector Algebra Q #875
KNOWLEDGE BASED
APPLY
2 Marks 2023 VSA
(a) If the vectors →a and →b are such that |→a|=3, |→b|=2/3 and →a x →b is a unit vector, then find the angle between →a and →b. OR (b) Find the area of a parallelogram whose adjacent sides are determined by the vectors →a = î - î + 3ê and →b = 2î - 7î + ê.
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Step-by-Step Solution

(a)

Given: |→a| = 3, |→b| = 2/3, |→a x →b| = 1

We know that |→a x →b| = |→a| |→b| sin θ, where θ is the angle between →a and →b.

So, 1 = 3 * (2/3) * sin θ

1 = 2 sin θ

sin θ = 1/2

θ = π/6

(b)

Given: →a = î - ĵ + 3k and →b = 2î - 7ĵ + k

Area of parallelogram = |→a x →b|

→a x →b = | î ĵ k |

| 1 -1 3 |

| 2 -7 1 |

→a x →b = î(-1 + 21) - ĵ(1 - 6) + k(-7 + 2)

→a x →b = 20î + 5ĵ - 5k

|→a x →b| = √(20² + 5² + (-5)²) = √(400 + 25 + 25) = √450 = 15√2

Correct Answer: (a) π/6 OR (b) 15√2

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires students to apply the formula for the magnitude of the cross product of two vectors to find the angle between them.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure to calculate the angle between two vectors or the area of a parallelogram. This involves using formulas and applying them in a step-by-step manner.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of vector algebra and its applications, as covered in the textbook.

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