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Quiz 1
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The position vector of the point (1, 0, 2) is
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\(\vec{i}\) +\(\vec{j}\) + 2\(\vec{k}\)
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\(\vec{i}\) + 2\(\vec{j}\)
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\(\vec{2}\) + 3\(\vec{k}\)
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\(\vec{i}\) + 2\(\vec{K}\)
Explanation
\(\vec{i}\) + 2\(\vec{K}\)
The modulus of 7\(\vec{i}\) – 2\(\vec{J}\) + \(\vec{K}\)
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\(\sqrt{10}\)
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\(\sqrt{55}\)
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3\(\sqrt{6}\)
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6
Explanation
3\(\sqrt{6}\)
If O be the origin and \(\vec{OP}\) = 2\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\) and \(\vec{OQ}\) = 5\(\hat{i}\) + 4\(\hat{j}\) -3\(\hat{k}\), then \(\vec{PQ}\) is equal to
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7\(\hat{i}\) + 7\(\hat{j}\) – 7\(\hat{k}\)
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-3\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
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-7\(\hat{i}\) – 7\(\hat{j}\) + 7\(\hat{k}\)
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3\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
Explanation
3\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
The scalar product of 5\(\hat{i}\) + \(\hat{j}\) – 3\(\hat{k}\) and 3\(\hat{i}\) – 4\(\hat{j}\) + 7\(\hat{k}\) is
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10
0%
-10
0%
15
0%
-15
Explanation
-10
If \(\vec{a}\).\(\vec{b}\) = 0, then
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a ⊥ b
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\(\vec{a}\) || \(\vec{b}\)
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\(\vec{a}\) + \(\vec{b}\) = 0
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\(\vec{a}\) – \(\vec{b}\) = 0
Explanation
a ⊥ b
\(\vec{i}\) – \(\vec{j}\) =
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0
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1
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\(\vec{k}\)
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–\(\vec{k}\)
Explanation
0
\(\vec{k}\) × \(\vec{j}\) =
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0
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1
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\(\vec{i}\)
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–\(\vec{i}\)
Explanation
–\(\vec{i}\)
\(\vec{a}\). \(\vec{a}\) =
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0
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1
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|\(\vec{a}\)|²
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|\(\vec{a}\)|
Explanation
|\(\vec{a}\)|²
The projection of the vector 2\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) on the vector \(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) is
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\(\frac{4}{√6}\)
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\(\frac{5}{√6}\)
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\(\frac{4}{√3}\)
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\(\frac{7}{√6}\)
Explanation
\(\frac{5}{√6}\)
If |\(\vec{a}\)|= \(\sqrt{26}\), |b| = 7 and |\(\vec{a}\) × \(\vec{b}\)| = 35, then \(\vec{a}\).\(\vec{b}\) =
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8
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7
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9
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12
Explanation
7
If \(\vec{a}\) = 2\(\vec{i}\) – 3\(\vec{j}\) + 4\(\vec{k}\) and \(\vec{b}\) = \(\vec{i}\) + 2\(\vec{j}\) + \(\vec{k}\) then \(\vec{a}\) + \(\vec{b}\) =
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\(\vec{i}\) + \(\vec{j}\) + 3\(\vec{k}\)
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3\(\vec{i}\) – \(\vec{j}\) + 5\(\vec{k}\)
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\(\vec{i}\) – \(\vec{j}\) – 3\(\vec{k}\)
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2\(\vec{i}\) + \(\vec{j}\) + \(\vec{k}\)
Explanation
3\(\vec{i}\) – \(\vec{j}\) + 5\(\vec{k}\)
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 =
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\(\frac{6}{7}\)
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\(\frac{5}{7}\)
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\(\frac{4}{7}\)
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\(\frac{1}{2}\)
Explanation
\(\frac{5}{7}\)
If |\(\vec{a}\) + \(\vec{b}\)| = |\(\vec{a}\) – \(\vec{b}\)|, then
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\(\vec{a}\) || \(\vec{a}\)
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\(\vec{a}\) ⊥ \(\vec{b}\)
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|\(\vec{a}\)| = |\(\vec{b}\)|
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None of these
Explanation
\(\vec{a}\) ⊥ \(\vec{b}\)
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
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\(\frac{2}{3}\)
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\(\frac{4}{3}\)
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–\(\frac{4}{3}\)
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\(\frac{5}{3}\)
Explanation
\(\frac{4}{3}\)
If |\(\vec{a}\) × \(\vec{b}\)| – |\(\vec{a}\).\(\vec{b}\)|, then the angle between \(\vec{a}\) and \(\vec{b}\), is
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0
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\(\frac{π}{2}\)
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\(\frac{π}{4}\)
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π
Explanation
\(\frac{π}{4}\)
The angle between two vector \(\vec{a}\) and \(\vec{b}\) with magnitude √3 and 4, respectively and \(\vec{a}\).\(\vec{b}\) = 2√3 is
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\(\frac{π}{6}\)
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\(\frac{π}{3}\)
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\(\frac{π}{2}\)
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\(\frac{5π}{2}\)
Explanation
\(\frac{π}{3}\)
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
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\(\frac{\hat{i}+\hat{j}+\hat{k}}{√3}\)
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\(\frac{\hat{i}-\hat{j}+\hat{k}}{√3}\)
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\(\frac{\hat{i}-\hat{j}-\hat{k}}{√3}\)
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\(\frac{\hat{i}+\hat{j}-\hat{k}}{√3}\)
Explanation
\(\frac{\hat{i}-\hat{j}-\hat{k}}{√3}\)
If \(\vec{a}\) = 2\(\vec{i}\) – 5\(\vec{j}\) + k and \(\vec{b}\) = 4\(\vec{i}\) + 2\(\vec{j}\) + \(\vec{k}\) then \(\vec{a}\).\(\vec{b}\) =
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0
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-1
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1
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2
Explanation
-1
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
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The A, B and C are collinear
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A, B and C are not colinear
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\(\vec{AB}\) ⊥ \(\vec{BC}\)
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None of these
Explanation
The A, B and C are collinear
According to the associative lass of addition of addition of s ector
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\(\vec{b}\), \(\vec{a}\)
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\(\vec{a}\), \(\vec{b}\)
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\(\vec{a}\), 0
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\(\vec{b}\), 0
Explanation
\(\vec{b}\), \(\vec{a}\)
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