CBSE  /  Class 12  /  Maths  /  Vector Algebra

  • 1. 
    If |\(\vec{a}\)|= \(\sqrt{26}\), |b| = 7 and |\(\vec{a}\) × \(\vec{b}\)| = 35, then \(\vec{a}\).\(\vec{b}\) =

  • 8
  • 7
  • 9
  • 12

  • 2. 
    If \(\vec{a}\) = 2\(\vec{i}\) – 3\(\vec{j}\) + 4\(\vec{k}\) and \(\vec{b}\) = \(\vec{i}\) + 2\(\vec{j}\) + \(\vec{k}\) then \(\vec{a}\) + \(\vec{b}\) =

  • \(\vec{i}\) + \(\vec{j}\) + 3\(\vec{k}\)
  • 3\(\vec{i}\) – \(\vec{j}\) + 5\(\vec{k}\)
  • \(\vec{i}\) – \(\vec{j}\) – 3\(\vec{k}\)
  • 2\(\vec{i}\) + \(\vec{j}\) + \(\vec{k}\)

  • 3. 
    If \(\vec{a}\) = \(\vec{i}\) + 2\(\vec{j}\) + 3\(\vec{k}\) and \(\vec{b}\) = 3\(\vec{i}\) + 2\(\vec{j}\) + \(\vec{k}\), then cos θ =

  • \(\frac{6}{7}\)
  • \(\frac{5}{7}\)
  • \(\frac{4}{7}\)
  • \(\frac{1}{2}\)

  • 4. 
    If |\(\vec{a}\) + \(\vec{b}\)| = |\(\vec{a}\) – \(\vec{b}\)|, then

  • \(\vec{a}\) || \(\vec{a}\)
  • \(\vec{a}\) ⊥ \(\vec{b}\)
  • |\(\vec{a}\)| = |\(\vec{b}\)|
  • None of these

  • 5. 
    The projection of the vector 2\(\hat{i}\) + 3\(\hat{j}\) – 6\(\hat{k}\) on the line joining the points (3, 4, 2) and (5, 6,3) is

  • \(\frac{2}{3}\)
  • \(\frac{4}{3}\)
  • –\(\frac{4}{3}\)
  • \(\frac{5}{3}\)

  • 6. 
    If |\(\vec{a}\) × \(\vec{b}\)| – |\(\vec{a}\).\(\vec{b}\)|, then the angle between \(\vec{a}\) and \(\vec{b}\), is

  • 0
  • \(\frac{π}{2}\)
  • \(\frac{π}{4}\)
  • π

  • 7. 
    The angle between two vector \(\vec{a}\) and \(\vec{b}\) with magnitude √3 and 4, respectively and \(\vec{a}\).\(\vec{b}\) = 2√3 is

  • \(\frac{π}{6}\)
  • \(\frac{π}{3}\)
  • \(\frac{π}{2}\)
  • \(\frac{5π}{2}\)

  • 8. 
    Unit vector perpendicular to each of the vector 3\(\hat{i}\) + \(\hat{j}\) + 2\(\hat{k}\) and 2\(\hat{i}\) – 2\(\hat{j}\) + 4\(\hat{k}\) is

  • \(\frac{\hat{i}+\hat{j}+\hat{k}}{√3}\)
  • \(\frac{\hat{i}-\hat{j}+\hat{k}}{√3}\)
  • \(\frac{\hat{i}-\hat{j}-\hat{k}}{√3}\)
  • \(\frac{\hat{i}+\hat{j}-\hat{k}}{√3}\)

  • 9. 
    If \(\vec{a}\) = 2\(\vec{i}\) – 5\(\vec{j}\) + k and \(\vec{b}\) = 4\(\vec{i}\) + 2\(\vec{j}\) + \(\vec{k}\) then \(\vec{a}\).\(\vec{b}\) =

  • 0
  • -1
  • 1
  • 2

  • 10. 
    If 2\(\vec{i}\) + \(\vec{j}\) + \(\vec{k}\), 6\(\vec{i}\) – \(\vec{j}\) + 2\(\vec{k}\) and 14\(\vec{i}\) – 5\(\vec{j}\) + 4\(\vec{k}\) be the position vector of the points A, B and C respectively, then

  • The A, B and C are collinear
  • A, B and C are not colinear
  • \(\vec{AB}\) ⊥ \(\vec{BC}\)
  • None of these
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