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Vector Algebra Questions and Answers

31.

The projection of \(\vec{b}\) on \(\vec{a}\) is 12. If the angle between \(\vec{a}\) and \(\vec{b}\) is 60°, then \(|\vec{b}|\)=

Exam Year(s)

KEAM - 2026,

Answer: E

\(|\vec{b}|\) = 24.

The scalar projection of \(\vec{b}\) on \(\vec{a}\) is \(|\vec{b}|\cos\theta\).

\[ \begin{aligned} |\vec{b}|\cos 60^{\circ} &= 12 \\ |\vec{b}| \times \tfrac{1}{2} &= 12 \\ |\vec{b}| &= 24 \end{aligned} \]

6 comes from multiplying 12 by \(\cos 60^{\circ}\) instead of dividing, and 12 treats the projection as the whole vector.

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32.

Let \(\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k}\) and \(\vec{b} = 2\hat{i} + \hat{j} + \alpha\hat{k}\). If \(|\vec{a} + \vec{b}| = \sqrt{29}\). Then the possible values of \(\alpha\) are

Exam Year(s)

KEAM - 2026,

Answer: C

\(\alpha\) = 3 or 7.

\[ \begin{aligned} \vec{a} + \vec{b} &= 4\hat{i} + 3\hat{j} + (\alpha - 5)\hat{k} \\ 16 + 9 + (\alpha - 5)^{2} &= 29 \\ (\alpha - 5)^{2} &= 4 \\ \alpha - 5 &= \pm 2 \end{aligned} \]

So \(\alpha = 7\) or \(\alpha = 3\). The answer is \(5 \pm 2\), not \(\pm 2\): forgetting the -5 already in \(\vec{a}\) gives option A, and changing the sign of 5 gives -3, -7.

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33.

If the position vectors of the points P and Q are, respectively, \(5\vec{a} - 6\vec{b}\) and \(\vec{a} + 2\vec{b}\), then the point R with position vector \(2\vec{a}\) divides the line segment joining P and Q internally in the ratio

Exam Year(s)

KEAM - 2026,

Answer: B

R divides PQ in the ratio 3:1.

If R divides PQ internally in the ratio m : n, then \(\vec{r} = \frac{m\vec{q} + n\vec{p}}{m + n}\).

\[ \begin{aligned} 2\vec{a}(m + n) &= m(\vec{a} + 2\vec{b}) + n(5\vec{a} - 6\vec{b}) \\ \vec{b}:\ 0 &= 2m - 6n \Rightarrow m = 3n \\ \vec{a}:\ 2(m + n) &= m + 5n \ \checkmark \end{aligned} \]

With m = 3, n = 1 the \(\vec{a}\) coefficients are 8 = 3 + 5, so the ratio is 3 : 1. 1:3 (not offered) would be the same division read from Q's end; 2:1 and 3:2 do not cancel the \(\vec{b}\) terms.

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34.

Let

\[ \begin{aligned} \vec{a} &= (\sin^{2}\alpha)\hat{i} + (\cos 2\alpha)\hat{j} \\ &\quad + (\cos^{2}\alpha)\hat{k},\ 0 \le \alpha \le \frac{\pi}{2} \end{aligned} \]

and \(\vec{b} = \hat{i} - 2\hat{j} + \hat{k}\). If \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other, then the value of \(\alpha\) is equal to

Exam Year(s)

KEAM - 2026,

Answer: A

\(\alpha\) = \(\frac{\pi}{6}\).

Perpendicular vectors have zero dot product:

\[ \begin{aligned} \sin^{2}\alpha - 2\cos 2\alpha + \cos^{2}\alpha &= 0 \\ 1 - 2\cos 2\alpha &= 0 \\ \cos 2\alpha &= \tfrac{1}{2} \end{aligned} \]

With \(0 \le \alpha \le \frac{\pi}{2}\), \(2\alpha = \frac{\pi}{3}\), so \(\alpha = \frac{\pi}{6}\). \(\frac{\pi}{4}\) would make \(\cos 2\alpha = 0\) and the dot product 1; \(\frac{\pi}{3}\) gives \(\cos 2\alpha = -\frac{1}{2}\) and a dot product of 2.

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35.

Let \(O\) be the origin. Let \(\overrightarrow{OA} = \vec{a}\) and \(\overrightarrow{OB} = \vec{b}\) be the position vectors of the points \(A\) and \(B\) respectively. A point \(P\) divides the line segment \(AB\) internally in the ratio \(m : n\) . Then \(\overrightarrow{AP}\) is equal to

Exam Year(s)

KEAM - 2026,

Answer: D

\(\overrightarrow{AP} = \frac{m(\vec{b} - \vec{a})}{m + n}\).

Since \(P\) divides \(AB\) internally in the ratio \(m : n\), \(AP\) is \(\frac{m}{m + n}\) of \(AB\), in the same direction:

\[ \begin{aligned} \overrightarrow{AP} &= \frac{m}{m + n}\,\overrightarrow{AB} \\ &= \frac{m(\vec{b} - \vec{a})}{m + n} \end{aligned} \]

Equivalently, \(\overrightarrow{OP} = \frac{m\vec{b} + n\vec{a}}{m + n}\), and subtracting \(\vec{a}\) gives the same result. \(\frac{n(\vec{b} - \vec{a})}{m + n}\) is \(\overrightarrow{PB}\), the other part of the segment, and the forms with \(\vec{b} + \vec{a}\) or \(m - n\) belong to other constructions.

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36.

If \(2\hat{i} - \hat{j} + \hat{k} = s(3\hat{i} - 4\hat{j} - 4\hat{k})\) \(+\, t(\hat{i} - 3\hat{j} - 5\hat{k})\) , where \(s\) and \(t\) are scalars, then \(3s + 5t\) is equal to

Exam Year(s)

KEAM - 2026,

Answer: C

\(3s + 5t = -2\).

Compare components:

\[ \begin{aligned} 3s + t &= 2 \\ -4s - 3t &= -1 \\ -4s - 5t &= 1 \end{aligned} \]

From the first, \(t = 2 - 3s\). Substituting in the second: \(-4s - 6 + 9s = -1\), so \(s = 1\) and \(t = -1\). The third checks: \(-4 + 5 = 1\).

\[ 3s + 5t = 3 - 5 = -2 \]

The value 2 is \(3s + t\), the first component, not the quantity asked.

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37.

Let \(\vec{a} = 2\hat{i} - 2\hat{j} + 4\hat{k}\) , \(\vec{b} = -5\hat{i} - \hat{j} + 8\hat{k}\) and \(\vec{c} = 3\hat{i} + \hat{j} - \lambda\hat{k}\) . If \(\vec{a} + \vec{b} + \vec{c}\) and \(\vec{a} - \vec{b} + \vec{c}\) are perpendicular, then the values of \(\lambda\) are

Exam Year(s)

KEAM - 2026,

Answer: E

\(\lambda = -4\) and \(12\).

\[ \begin{aligned} \vec{a} + \vec{b} + \vec{c} &= 0\hat{i} - 2\hat{j} + (12 - \lambda)\hat{k} \\ \vec{a} - \vec{b} + \vec{c} &= 10\hat{i} + 0\hat{j} + (-4 - \lambda)\hat{k} \end{aligned} \]

Perpendicular vectors have zero dot product:

\[ \begin{aligned} 0 + 0 + (12 - \lambda)(-4 - \lambda) &= 0 \\ \lambda &= 12 \text{ or } \lambda = -4 \end{aligned} \]

The option 4 and -12 has both signs reversed, and the other pairs do not make either factor zero.

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38.

If \(|\vec{a}| = \sqrt{26}\), \(|\vec{b}| = \sqrt{3}\) and \(\vec{a} \times \vec{b} = 5\hat{i} + \hat{j} - 4\hat{k}\) ,then \(\vec{a} \cdot \vec{b} =\)

Exam Year(s)

KEAM - 2026,

Answer: E

\(\vec{a} \cdot \vec{b} = \pm 6\).

Use Lagrange's identity, \(|\vec{a} \times \vec{b}|^{2} + (\vec{a} \cdot \vec{b})^{2}\) \(= |\vec{a}|^{2}|\vec{b}|^{2}\):

\[ \begin{aligned} |\vec{a} \times \vec{b}|^{2} &= 25 + 1 + 16 = 42 \\ (\vec{a} \cdot \vec{b})^{2} &= 26 \times 3 - 42 = 36 \\ \vec{a} \cdot \vec{b} &= \pm 6 \end{aligned} \]

The cross product fixes only the magnitude of the dot product, not its sign, hence \(\pm\). The other options do not satisfy the identity.

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39.

The vectors \(2\vec{a} + 6\vec{b}\) and \(3\vec{a} - 7\vec{b}\) are position vectors of the points \(A\) and \(B\) respectively. A point \(P\) divides the line segment \(AB\) internally in the ratio 3:5 . Then \(\vec{PB}\) =

Exam Year(s)

KEAM - 2026,

Answer: A

\(\vec{PB}\) is \(\dfrac{5\vec{a} - 65\vec{b}}{8}\).

By the section formula, P dividing AB internally in the ratio 3 : 5 has position vector

\[ \begin{aligned} \vec{P} &= \frac{5(2\vec{a} + 6\vec{b}) + 3(3\vec{a} - 7\vec{b})}{8} \\ &= \frac{19\vec{a} + 9\vec{b}}{8} \end{aligned} \]

\[ \begin{aligned} \vec{PB} &= \vec{B} - \vec{P} \\ &= \frac{24\vec{a} - 56\vec{b} - 19\vec{a} - 9\vec{b}}{8} \\ &= \frac{5\vec{a} - 65\vec{b}}{8} \end{aligned} \]

Check: \(\vec{PB}\) should be \(\frac{5}{8}\) of \(\vec{AB} = \vec{a} - 13\vec{b}\), and it is. The other options get the \(\vec{b}\) coefficient or its sign wrong.

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40.

The position vectors of the vertices of a triangle are \(\hat{i} + 2\hat{j} - 4\hat{k}\) , \(-\hat{i} - 2\hat{j} - 4\hat{k}\) and \(2\hat{i} + 3\hat{j} + 5\hat{k}\) . The perimeter of the triangle is

Exam Year(s)

KEAM - 2026,

Answer: A

The perimeter is \(\sqrt{20} + \sqrt{115} + \sqrt{83}\).

With vertices \(A(1, 2, -4)\), \(B(-1, -2, -4)\) and \(C(2, 3, 5)\):

\[ \begin{aligned} AB &= \sqrt{4 + 16 + 0} = \sqrt{20} \\ BC &= \sqrt{9 + 25 + 81} = \sqrt{115} \\ CA &= \sqrt{1 + 1 + 81} = \sqrt{83} \end{aligned} \]

The perimeter is the sum of the three. Each wrong option changes one of the three square roots by a small amount, as a single arithmetic slip in squaring a difference would.

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Tags: Vector Algebra Questions and Answers