The vectors \(\vec{a}\) = x\(\hat{i}\) – 2\(\hat{j}\) + 5\(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + y\(\hat{j}\) – z\(\hat{k}\) are collinear, if
  • x = 1, y = -2, z = -5
  • x = \(\frac{3}{2}\), y = -4, z = -10
  • x = \(\frac{3}{2}\), y = 4, z = 10
  • All of these
The vectors (x, x + 1, x + 2), (x + 3, x + 4, x + 5) and (x + 6, x + 7, x + 8) are coplanar for
  • all values of x
  • x < 0
  • x ≤ 0
  • None of these
The vectors \(\vec{AB}\) = 3\(\hat{i}\) +4\(\hat{k}\) and \(\vec{AC}\) = 5\(\hat{i}\) – 2\(\hat{j}\) + 4\(\hat{k}\) are the sides of ΔABC. The length of the median through A is
  • \(\sqrt{18}\)
  • \(\sqrt{72}\)
  • \(\sqrt{33}\)
  • \(\sqrt{288}\)
The summation of two unit vectors is a third unit vector, then the modulus of the difference of the unit vector is
  • √3
  • 1 – √3
  • 1 + √3
  • -√3
0 h : 0 m : 1 s

Answered Not Answered Not Visited Correct : 0 Incorrect : 0