Every multiple-choice question from Applications of Vector Algebra (12th Standard Mathematics, Samacheer Kalvi) with the correct option highlighted and a clear, worked explanation. 25 questions in all — free to read in English and Tamil.
Q1
If \(\vec{a}\) and \(\vec{b}\) are parallel vectors, then \([\vec{a},\vec{c},\vec{b}]\) is equal to:
- A. \(2\)
- B. \(-1\)
- C. \(1\)
- D. \(0\)Correct
Explanation. Parallel vectors satisfy \(\vec{a}\times\vec{b}=\vec{0}\). Interchanging two entries of a scalar triple product only flips its sign, so \([\vec{a},\vec{c},\vec{b}]=-[\vec{a},\vec{b},\vec{c}]=-(\vec{a}\times\vec{b})\cdot\vec{c}\). Because \(\vec{a}\times\vec{b}=\vec{0}\), this equals \(0\). Equivalently, three vectors of which two are parallel are coplanar, and the box product of coplanar vectors vanishes.
Q2
If a vector \(\vec{\alpha}\) lies in the plane of \(\vec{\beta}\) and \(\vec{\gamma}\), then:
- A. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=1\)
- B. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=-1\)
- C. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\)Correct
- D. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=2\)
Explanation. Any vector lying in the plane of \(\vec{\beta}\) and \(\vec{\gamma}\) can be written as \(\vec{\alpha}=s\vec{\beta}+t\vec{\gamma}\). The three vectors are then coplanar, and the scalar triple product of coplanar vectors is always zero, so \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\). Geometrically the box product is the volume of the parallelepiped on the three vectors, which collapses to zero when they share a plane.
Q3
If \(\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=0\), then the value of \(\big|[\vec{a},\vec{b},\vec{c}]\big|\) is:
- A. \(|\vec{a}|\,|\vec{b}|\,|\vec{c}|\)Correct
- B. \(\dfrac{1}{3}|\vec{a}|\,|\vec{b}|\,|\vec{c}|\)
- C. \(1\)
- D. \(-1\)
Explanation. The three conditions mean the vectors are mutually perpendicular. For an orthogonal triad, \(\vec{b}\times\vec{c}\) is aligned with \(\vec{a}\) and has magnitude \(|\vec{b}|\,|\vec{c}|\). Hence \(\big|[\vec{a},\vec{b},\vec{c}]\big|=\big|\vec{a}\cdot(\vec{b}\times\vec{c})\big|=|\vec{a}|\,|\vec{b}|\,|\vec{c}|\), the volume of the rectangular box with the three vectors as edges.
Q4
If \(\vec{a},\vec{b},\vec{c}\) are three unit vectors such that \(\vec{a}\) is perpendicular to \(\vec{b}\) and is parallel to \(\vec{c}\), then \(\vec{a}\times(\vec{b}\times\vec{c})\) is equal to:
- A. \(\vec{a}\)
- B. \(\vec{b}\)Correct
- C. \(\vec{c}\)
- D. \(\vec{0}\)
Explanation. Apply the vector triple product identity \(\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\). Since \(\vec{a}\parallel\vec{c}\) and both are unit vectors, \(\vec{a}\cdot\vec{c}=1\); since \(\vec{a}\perp\vec{b}\), \(\vec{a}\cdot\vec{b}=0\). The expression collapses to \(1\cdot\vec{b}-0\cdot\vec{c}=\vec{b}\).
Q5
If \([\vec{a},\vec{b},\vec{c}]=1\), then the value of \(\dfrac{\vec{a}\cdot(\vec{b}\times\vec{c})}{(\vec{c}\times\vec{a})\cdot\vec{b}}+\dfrac{\vec{b}\cdot(\vec{c}\times\vec{a})}{(\vec{a}\times\vec{b})\cdot\vec{c}}+\dfrac{\vec{c}\cdot(\vec{a}\times\vec{b})}{(\vec{c}\times\vec{b})\cdot\vec{a}}\) is:
- A. \(1\)Correct
- B. \(-1\)
- C. \(2\)
- D. \(3\)
Explanation. Each numerator is a cyclic rearrangement of \([\vec{a},\vec{b},\vec{c}]\), so each equals \(1\). Among the denominators, \((\vec{c}\times\vec{a})\cdot\vec{b}=[\vec{c},\vec{a},\vec{b}]=1\) and \((\vec{a}\times\vec{b})\cdot\vec{c}=[\vec{a},\vec{b},\vec{c}]=1\) are cyclic permutations, but \((\vec{c}\times\vec{b})\cdot\vec{a}=[\vec{c},\vec{b},\vec{a}]=-1\) is an odd permutation. The sum is therefore \(\tfrac{1}{1}+\tfrac{1}{1}+\tfrac{1}{-1}=1\).
Q6
The volume of the parallelepiped whose edges are represented by the vectors \(\hat{i}+\hat{j},\ \hat{i}+2\hat{j},\ \hat{i}+\hat{j}+\pi\hat{k}\) is:
- A. \(\dfrac{\pi}{2}\)
- B. \(\dfrac{\pi}{3}\)
- C. \(\pi\)Correct
- D. \(\dfrac{\pi}{4}\)
Explanation. The volume is the magnitude of the determinant whose rows are the three edge vectors: \[\begin{vmatrix}1&1&0\\1&2&0\\1&1&\pi\end{vmatrix}.\] Expanding along the third column, only the entry \(\pi\) contributes, giving \(\pi\,(1\cdot2-1\cdot1)=\pi\). The volume is \(\pi\) cubic units.
Q7
If \(\vec{a}\) and \(\vec{b}\) are unit vectors such that \([\vec{a},\vec{b},\vec{a}\times\vec{b}]=\dfrac{1}{4}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is:
- A. \(\dfrac{\pi}{6}\)Correct
- B. \(\dfrac{\pi}{4}\)
- C. \(\dfrac{\pi}{3}\)
- D. \(\dfrac{\pi}{2}\)
Explanation. Here \([\vec{a},\vec{b},\vec{a}\times\vec{b}]=(\vec{a}\times\vec{b})\cdot(\vec{a}\times\vec{b})=|\vec{a}\times\vec{b}|^{2}\). For unit vectors \(|\vec{a}\times\vec{b}|=\sin\theta\), so the box product equals \(\sin^{2}\theta\). Setting \(\sin^{2}\theta=\tfrac14\) gives \(\sin\theta=\tfrac12\), hence \(\theta=\tfrac{\pi}{6}\).
Q8
If \(\vec{a}=\hat{i}+\hat{j}+\hat{k},\ \vec{b}=\hat{i}+\hat{j},\ \vec{c}=\hat{i}\) and \((\vec{a}\times\vec{b})\times\vec{c}=\lambda\vec{a}+\mu\vec{b}\), then the value of \(\lambda+\mu\) is:
- A. \(0\)Correct
- B. \(1\)
- C. \(6\)
- D. \(3\)
Explanation. First \(\vec{a}\times\vec{b}=-\hat{i}+\hat{j}\). Crossing with \(\vec{c}=\hat{i}\) gives \((-\hat{i}+\hat{j})\times\hat{i}=-\hat{k}\). Writing \(\lambda\vec{a}+\mu\vec{b}=(\lambda+\mu)\hat{i}+(\lambda+\mu)\hat{j}+\lambda\hat{k}\) and matching with \(-\hat{k}\): the \(\hat{k}\) term gives \(\lambda=-1\) and the \(\hat{i}\) term gives \(\lambda+\mu=0\), so \(\mu=1\). Therefore \(\lambda+\mu=0\).
Q9
If \(\vec{a},\vec{b},\vec{c}\) are non-coplanar, non-zero vectors such that \([\vec{a},\vec{b},\vec{c}]=3\), then \(\big\{[\vec{a}\times\vec{b},\ \vec{b}\times\vec{c},\ \vec{c}\times\vec{a}]\big\}^{2}\) is equal to:
- A. \(81\)Correct
- B. \(9\)
- C. \(27\)
- D. \(18\)
Explanation. A standard identity gives \([\vec{a}\times\vec{b},\ \vec{b}\times\vec{c},\ \vec{c}\times\vec{a}]=[\vec{a},\vec{b},\vec{c}]^{2}\). With \([\vec{a},\vec{b},\vec{c}]=3\), the inner box product equals \(3^{2}=9\). Squaring as the question asks gives \(9^{2}=81\).
Q10
If \(\vec{a},\vec{b},\vec{c}\) are three non-coplanar unit vectors such that \(\vec{a}\times(\vec{b}\times\vec{c})=\dfrac{\vec{b}+\vec{c}}{\sqrt{2}}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is:
- A. \(\dfrac{\pi}{2}\)
- B. \(\dfrac{3\pi}{4}\)Correct
- C. \(\dfrac{\pi}{4}\)
- D. \(\pi\)
Explanation. Expanding the left side, \(\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\). Comparing with \(\dfrac{\vec{b}+\vec{c}}{\sqrt{2}}\) and using that \(\vec{b},\vec{c}\) are independent, \(\vec{a}\cdot\vec{c}=\tfrac{1}{\sqrt2}\) and \(\vec{a}\cdot\vec{b}=-\tfrac{1}{\sqrt2}\). Since all are unit vectors, \(\cos\theta=\vec{a}\cdot\vec{b}=-\tfrac{1}{\sqrt2}\), giving \(\theta=\tfrac{3\pi}{4}\).
Q11
If the volume of the parallelepiped with \(\vec{a}\times\vec{b},\ \vec{b}\times\vec{c},\ \vec{c}\times\vec{a}\) as coterminous edges is \(8\) cubic units, then the volume of the parallelepiped with \((\vec{a}\times\vec{b})\times(\vec{b}\times\vec{c}),\ (\vec{b}\times\vec{c})\times(\vec{c}\times\vec{a})\) and \((\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b})\) as coterminous edges is:
- A. \(8\) cubic units
- B. \(512\) cubic units
- C. \(64\) cubic unitsCorrect
- D. \(24\) cubic units
Explanation. Write \(V=[\vec{a},\vec{b},\vec{c}]\). The first volume satisfies \([\vec{a}\times\vec{b},\vec{b}\times\vec{c},\vec{c}\times\vec{a}]=V^{2}=8\). For the second set, the identity \([(\vec{a}\times\vec{b})\times(\vec{b}\times\vec{c}),\ (\vec{b}\times\vec{c})\times(\vec{c}\times\vec{a}),\ (\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b})]=[\vec{a}\times\vec{b},\vec{b}\times\vec{c},\vec{c}\times\vec{a}]^{2}=V^{4}\) applies. Hence the volume is \((V^{2})^{2}=8^{2}=64\) cubic units.
Q12
Consider the vectors \(\vec{a},\vec{b},\vec{c},\vec{d}\) such that \((\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d})=\vec{0}\). Let \(P_{1}\) and \(P_{2}\) be the planes determined by the pairs \(\{\vec{a},\vec{b}\}\) and \(\{\vec{c},\vec{d}\}\) respectively. Then the angle between \(P_{1}\) and \(P_{2}\) is:
- A. \(0\)Correct
- B. \(45^{\circ}\)
- C. \(60^{\circ}\)
- D. \(90^{\circ}\)
Explanation. The vector \(\vec{a}\times\vec{b}\) is normal to \(P_{1}\) and \(\vec{c}\times\vec{d}\) is normal to \(P_{2}\). The condition \((\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d})=\vec{0}\) means these two normals are parallel. Planes whose normals are parallel are themselves parallel, so the angle between them is \(0\).
Q13
If \(\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\times\vec{b})\times\vec{c}\), where \(\vec{a},\vec{b},\vec{c}\) are any three vectors such that \(\vec{b}\cdot\vec{c}\neq0\) and \(\vec{a}\cdot\vec{b}\neq0\), then \(\vec{a}\) and \(\vec{c}\) are:
- A. perpendicular
- B. parallelCorrect
- C. inclined at an angle \(\dfrac{\pi}{3}\)
- D. inclined at an angle \(\dfrac{\pi}{6}\)
Explanation. Expand both sides. The left gives \((\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\); the right gives \((\vec{a}\cdot\vec{c})\vec{b}-(\vec{b}\cdot\vec{c})\vec{a}\). Cancelling the common term \((\vec{a}\cdot\vec{c})\vec{b}\) leaves \((\vec{a}\cdot\vec{b})\vec{c}=(\vec{b}\cdot\vec{c})\vec{a}\). Because \(\vec{a}\cdot\vec{b}\neq0\), this forces \(\vec{c}=\dfrac{\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{b}}\,\vec{a}\), a nonzero scalar multiple of \(\vec{a}\). Thus \(\vec{a}\) and \(\vec{c}\) are parallel.
Q14
If \(\vec{a}=2\hat{i}+3\hat{j}-\hat{k},\ \vec{b}=\hat{i}+2\hat{j}-5\hat{k},\ \vec{c}=3\hat{i}+5\hat{j}-\hat{k}\), then a vector perpendicular to \(\vec{a}\) and lying in the plane containing \(\vec{b}\) and \(\vec{c}\) is:
- A. \(-17\hat{i}+21\hat{j}-97\hat{k}\)
- B. \(17\hat{i}+21\hat{j}-123\hat{k}\)
- C. \(-17\hat{i}-21\hat{j}+97\hat{k}\)
- D. \(-17\hat{i}-21\hat{j}-97\hat{k}\)Correct
Explanation. A vector perpendicular to \(\vec{a}\) but lying in the plane of \(\vec{b}\) and \(\vec{c}\) is \(\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\). Here \(\vec{a}\cdot\vec{c}=6+15+1=22\) and \(\vec{a}\cdot\vec{b}=2+6+5=13\), so the vector is \(22\vec{b}-13\vec{c}\). Componentwise, \(22(\hat{i}+2\hat{j}-5\hat{k})-13(3\hat{i}+5\hat{j}-\hat{k})=-17\hat{i}-21\hat{j}-97\hat{k}\).
Q15
The angle between the lines \(\dfrac{x-2}{3}=\dfrac{y+1}{-2},\ z=2\) and \(\dfrac{x-1}{1}=\dfrac{2y+3}{3}=\dfrac{z+5}{2}\) is:
- A. \(\dfrac{\pi}{6}\)
- B. \(\dfrac{\pi}{4}\)
- C. \(\dfrac{\pi}{3}\)
- D. \(\dfrac{\pi}{2}\)Correct
Explanation. The first line has direction ratios \((3,-2,0)\) because \(z=2\) is constant. For the second, rewriting \(\dfrac{2y+3}{3}=\dfrac{y+3/2}{3/2}\) gives direction ratios proportional to \((1,\tfrac32,2)\), i.e. \((2,3,4)\). Their dot product is \(3\cdot2+(-2)\cdot3+0\cdot4=0\), so the lines are perpendicular and the angle is \(\tfrac{\pi}{2}\).
Q16
If the line \(\dfrac{x-2}{3}=\dfrac{y-1}{-5}=\dfrac{z+2}{2}\) lies in the plane \(x+3y-\alpha z+\beta=0\), then \((\alpha,\beta)\) is:
- A. \((-5,5)\)
- B. \((-6,7)\)Correct
- C. \((5,-5)\)
- D. \((6,-7)\)
Explanation. For the line to lie in the plane, its direction \((3,-5,2)\) must be perpendicular to the plane's normal \((1,3,-\alpha)\): \(3-15-2\alpha=0\Rightarrow\alpha=-6\). The point \((2,1,-2)\) on the line must also satisfy the plane: \(2+3(1)-\alpha(-2)+\beta=0\). Substituting \(\alpha=-6\) gives \(5-12+\beta=0\), so \(\beta=7\). Hence \((\alpha,\beta)=(-6,7)\).
Q17
The angle between the line \(\vec{r}=(\hat{i}+2\hat{j}-3\hat{k})+t(2\hat{i}+\hat{j}-2\hat{k})\) and the plane \(\vec{r}\cdot(\hat{i}+\hat{j})+4=0\) is:
- A. \(0^{\circ}\)
- B. \(30^{\circ}\)
- C. \(45^{\circ}\)Correct
- D. \(90^{\circ}\)
Explanation. For a line with direction \(\vec{b}=2\hat{i}+\hat{j}-2\hat{k}\) and a plane with normal \(\vec{n}=\hat{i}+\hat{j}\), the angle satisfies \(\sin\theta=\dfrac{|\vec{b}\cdot\vec{n}|}{|\vec{b}|\,|\vec{n}|}\). Here \(\vec{b}\cdot\vec{n}=2+1=3\), \(|\vec{b}|=3\) and \(|\vec{n}|=\sqrt2\), so \(\sin\theta=\dfrac{3}{3\sqrt2}=\dfrac{1}{\sqrt2}\), giving \(\theta=45^{\circ}\).
Q18
The coordinates of the point where the line \(\vec{r}=(6\hat{i}-\hat{j}-3\hat{k})+t(-\hat{i}+4\hat{k})\) meets the plane \(\vec{r}\cdot(\hat{i}+\hat{j}-\hat{k})=3\) are:
- A. \((2,1,0)\)
- B. \((7,-1,-7)\)
- C. \((1,2,-6)\)
- D. \((5,-1,1)\)Correct
Explanation. A general point on the line is \((6-t,\,-1,\,-3+4t)\). Substituting into the plane \(x+y-z=3\): \((6-t)+(-1)-(-3+4t)=3\), which simplifies to \(8-5t=3\), so \(t=1\). The intersection point is \((5,-1,1)\).
Q19
The distance from the origin to the plane \(3x+6y+2z+7=0\) is:
- A. \(0\)
- B. \(1\)Correct
- C. \(2\)
- D. \(3\)
Explanation. The distance from the origin to the plane \(ax+by+cz+d=0\) is \(\dfrac{|d|}{\sqrt{a^{2}+b^{2}+c^{2}}}\). Here it equals \(\dfrac{|7|}{\sqrt{9+36+4}}=\dfrac{7}{\sqrt{49}}=1\).
Q20
The distance between the planes \(x+2y+3z+7=0\) and \(2x+4y+6z+7=0\) is:
- A. \(\dfrac{\sqrt{7}}{2\sqrt{2}}\)Correct
- B. \(\dfrac{7}{2}\)
- C. \(\dfrac{\sqrt{7}}{2}\)
- D. \(\dfrac{7}{2\sqrt{2}}\)
Explanation. Scale the first plane to share the second's normal: \(2x+4y+6z+14=0\). The two parallel planes \(2x+4y+6z+14=0\) and \(2x+4y+6z+7=0\) are separated by \(\dfrac{|14-7|}{\sqrt{4+16+36}}=\dfrac{7}{\sqrt{56}}=\dfrac{7}{2\sqrt{14}}\). Rationalising, \(\dfrac{7}{2\sqrt{14}}=\dfrac{\sqrt{14}}{4}=\dfrac{\sqrt{7}}{2\sqrt{2}}\).
Q21
If the direction cosines of a line are \(\dfrac{1}{c},\dfrac{1}{c},\dfrac{1}{c}\), then:
- A. \(c=\pm3\)
- B. \(c=\pm\sqrt{3}\)Correct
- C. \(c>0\)
- D. \(0
Explanation. Direction cosines satisfy \(l^{2}+m^{2}+n^{2}=1\). Substituting \(\dfrac{1}{c^{2}}+\dfrac{1}{c^{2}}+\dfrac{1}{c^{2}}=1\) gives \(\dfrac{3}{c^{2}}=1\), so \(c^{2}=3\) and \(c=\pm\sqrt{3}\).
Q22
The vector equation \(\vec{r}=(\hat{i}-2\hat{j}-\hat{k})+t(6\hat{j}-\hat{k})\) represents a straight line passing through the points:
- A. \((0,6,-1)\) and \((1,-2,-1)\)
- B. \((0,6,-1)\) and \((-1,-4,-2)\)
- C. \((1,-2,-1)\) and \((1,4,-2)\)Correct
- D. \((1,-2,-1)\) and \((0,-6,1)\)
Explanation. Setting \(t=0\) gives the point \((1,-2,-1)\). Setting \(t=1\) gives \((1,\,-2+6,\,-1-1)=(1,4,-2)\). Both points have \(x=1\) because the direction vector has no \(\hat{i}\) component, so the line passes through \((1,-2,-1)\) and \((1,4,-2)\).
Q23
If the distance of the point \((1,1,1)\) from the origin is half of its distance from the plane \(x+y+z+k=0\), then the values of \(k\) are:
- A. \(\pm3\)
- B. \(\pm6\)
- C. \(-3,\ 9\)
- D. \(3,\ -9\)Correct
Explanation. The distance from \((1,1,1)\) to the origin is \(\sqrt3\). Its distance to the plane is \(\dfrac{|1+1+1+k|}{\sqrt3}=\dfrac{|3+k|}{\sqrt3}\). The condition \(\sqrt3=\tfrac12\cdot\dfrac{|3+k|}{\sqrt3}\) gives \(|3+k|=6\), so \(3+k=\pm6\), yielding \(k=3\) or \(k=-9\).
Q24
If the planes \(\vec{r}\cdot(2\hat{i}-\lambda\hat{j}+\hat{k})=3\) and \(\vec{r}\cdot(4\hat{i}+\hat{j}-\mu\hat{k})=5\) are parallel, then the values of \(\lambda\) and \(\mu\) are:
- A. \(\dfrac{1}{2},\ -2\)
- B. \(-\dfrac{1}{2},\ 2\)
- C. \(-\dfrac{1}{2},\ -2\)Correct
- D. \(\dfrac{1}{2},\ 2\)
Explanation. Parallel planes have proportional normals: \(\dfrac{2}{4}=\dfrac{-\lambda}{1}=\dfrac{1}{-\mu}\). The common ratio is \(\tfrac12\). From \(-\lambda=\tfrac12\), \(\lambda=-\tfrac12\); from \(\dfrac{1}{-\mu}=\tfrac12\), \(-\mu=2\), so \(\mu=-2\). Hence \(\lambda=-\tfrac12,\ \mu=-2\).
Q25
If the length of the perpendicular from the origin to the plane \(2x+3y+\lambda z=1\) (with \(\lambda>0\)) is \(\dfrac{1}{5}\), then the value of \(\lambda\) is:
- A. \(2\sqrt{3}\)Correct
- B. \(3\sqrt{2}\)
- C. \(0\)
- D. \(1\)
Explanation. Write the plane as \(2x+3y+\lambda z-1=0\). The perpendicular distance from the origin is \(\dfrac{|-1|}{\sqrt{4+9+\lambda^{2}}}=\dfrac{1}{\sqrt{13+\lambda^{2}}}\). Setting this equal to \(\tfrac15\) gives \(\sqrt{13+\lambda^{2}}=5\), so \(13+\lambda^{2}=25\) and \(\lambda^{2}=12\). Since \(\lambda>0\), \(\lambda=2\sqrt{3}\).