Chapter test · Grade 12 · Mathematics
Mathematics Part-II · Ch 4
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40 minutes · 20 questions
The magnitude of vector \(\vec{a} = 3\hat{i} + 4\hat{j}\) is:
The vector \(\hat{i} \times \hat{j}\) equals:
If \(\vec{a} = 2\hat{i} - \hat{j}\) and \(\vec{b} = \hat{i} + 3\hat{j}\), then \(\vec{a} + \vec{b}\) is:
If \(\vec{a} \cdot \vec{b} = 0\) and both are non-zero, then the angle between them is:
The unit vector in direction of \(3\hat{i} + 4\hat{j}\) is:
If \(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\) and \(\vec{b} = 2\hat{i} - \hat{j} + \hat{k}\), then \(\vec{a} \cdot \vec{b}\) is:
Cross product of \(\hat{i}\) and \(\hat{k}\) is:
Area of parallelogram with adjacent sides \(\vec{a} = \hat{i} + \hat{j}\) and \(\vec{b} = \hat{j} + \hat{k}\) is:
If vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular unit vectors, then \(|\vec{a} + \vec{b} + \vec{c}|\) is:
Position vector of midpoint of segment joining \(A(\vec{a})\) and \(B(\vec{b})\) is:
Assertion (A): The vectors \(2\hat{i} + 3\hat{j} - \hat{k}\) and \(-4\hat{i} - 6\hat{j} - 2\hat{k}\) are collinear.
Reason (R): Two vectors \(\vec{a}\) and \(\vec{b}\) are collinear if and only if \(\vec{b} = \lambda\vec{a}\) for some non-zero scalar \(\lambda\).
Assertion (A): If the points A and B are \((1, 2, -3)\) and \((3, -4, 0)\), then \(|\overrightarrow{AB}| = 7\).
Reason (R): \(\overrightarrow{AB}\) is found by subtracting the position vector of B from the position vector of A.
Assertion (A): The direction cosines of the vector \(\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}\) are \(\dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3}\).
Reason (R): The sum of the squares of the direction cosines of any vector is \(1\).
Which vector statements are correct?
Which statements are correct?
Match each expression with what it represents (\(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\)):
For unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) along axes, match:
A drone flies in a straight line from its launch point P to its landing point Q. Relative to a fixed origin, their position vectors (in metres) are \(\vec{p} = 2\hat{i} - \hat{j} + 3\hat{k}\) and \(\vec{q} = 5\hat{i} + 5\hat{j} - 3\hat{k}\). The drone takes a photo at the point R that divides PQ internally in the ratio 2 : 1 (PR : RQ = 2 : 1). The position vector of R is:
Find \(\lambda\) such that \(\vec{a} = \hat{i} + 2\hat{j} + \lambda\hat{k}\) and \(\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}\) are perpendicular.
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