Q.Find $|\vec{a} \times \vec{b}|$, if $\vec{a}=\hat{i}-7\hat{j}+7\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+2\hat{k}$.

Showing the 12 most recent of 20 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If the vectors $\bar{A} = a\bar{i} + b\bar{j} + c\bar{k}$, $\bar{B} = d\bar{i} + 3\bar{j} + 4\bar{k}$, $\bar{C} = 3\bar{i} + \bar{j} - 2\bar{k}$ are such that $\bar{A} = \bar{B} + \bar{C}$ and form a triangle whose area is $5\sqrt{6}$ sq units, then the maximum value of $|a| + |b| + |c| + |d|$ is (A) $25$ (B) $27$ (C) $30$ (D) $33$

   ›Reveal solutionSolution

   Match components from $\bar A=\bar B+\bar C$, use the cross product to encode the triangle's area, solve a quadratic in $d$, and pick the branch giving the larger value.

   Concept and Intuition

   Two vectors $\bar B,\bar C$ placed tail-to-tail (or head-to-tail with resultant $\bar A$) form a triangle whose area is $\tfrac12|\bar B\times\bar C|$ — half the parallelogram area, exactly as in plane geometry. Given a numeric area, we get an equation in the unknown $d$, which is generally quadratic and gives two valid geometric configurations; the question asks for the larger of the resulting sums.

   Step-by-Step Solution

   1. $\bar A=\bar B+\bar C=(d+3)\bar i+(3+1)\bar j+(4-2)\bar k=(d+3)\bar i+4\bar j+2\bar k$. So $a=d+3,\,b=4,\,c=2$.
   2. Compute $\bar B\times\bar C$ with $\bar B=(d,3,4)$, $\bar C=(3,1,-2)$: $\bar B\times\bar C=\big(3(-2)-4(1),\ -(d(-2)-4(3)),\ d(1)-3(3)\big)=(-10,\ 2d+12,\ d-9)$.
   3. Area $=\tfrac12|\bar B\times\bar C|=5\sqrt6 \Rightarrow |\bar B\times\bar C|=10\sqrt6\Rightarrow|\bar B\times\bar C|^2=600$.
   4. $100+(2d+12)^2+(d-9)^2=600$.
   5. Expand: $(2d+12)^2=4d^2+48d+144$, $(d-9)^2=d^2-18d+81$. Sum with the $100$: $5d^2+30d+325=600\Rightarrow 5d^2+30d-275=0\Rightarrow d^2+6d-55=0$.
   6. Solve: $d=\dfrac{-6\pm\sqrt{36+220}}{2}=\dfrac{-6\pm16}{2}=5$ or $-11$.
   7. $|a|+|b|+|c|+|d|=|d+3|+4+2+|d|$.
      • $d=5$: $|8|+4+2+5=19$.
      • $d=-11$: $|-8|+4+2+11=8+4+2+11=25$.
   8. The maximum over both valid solutions is $25$.

   Common Mistakes

   • Only solving for one root of the quadratic and missing the branch that gives the maximum.
   • Sign slip in the $j$-component of the cross product ($-(\,)$ easy to drop).

   ✓Final answer

   The correct option is (A) — $25$.

   ANSWER: A

2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.In $\Delta ABC$, if $\overline{AB} = \bar{i} + \alpha\bar{j} + 2\bar{k}$, $\overline{BC} = \beta\bar{i} - 2\bar{j} + 3\bar{k}$ and $\overline{CA} = 2\bar{i} + 3\bar{j} - \gamma\bar{k}$, then the area of $\Delta ABC$ is (A) $\dfrac{1}{2}\sqrt{83}$ (B) $\dfrac{1}{2}\sqrt{107}$ (C) $\dfrac{1}{2}\sqrt{11}$ (D) $\dfrac{1}{2}\sqrt{22}$

   ›Reveal solutionSolution

   The triangle-closure condition $\overline{AB}+\overline{BC}+\overline{CA}=\bar0$ fixes the unknown scalars, then the area is half the magnitude of $\overline{AB}\times\overline{BC}$. Answer: $\frac12\sqrt{107}$.

   Concept and Intuition

   Going around a triangle $A\to B\to C\to A$ returns you to the start, so the three edge vectors (in that head-to-tail order) must sum to zero. This pins down $\alpha,\beta,\gamma$ without needing the vertices' actual coordinates. Once two edge vectors are known, the triangle's area is half the magnitude of their cross product.

   Step-by-Step Solution

   1. Closure: $\overline{AB}+\overline{BC}+\overline{CA}=\bar 0$.
   2. $\bar i$-component: $1+\beta+2=0 \Rightarrow \beta=-3$.
   3. $\bar j$-component: $\alpha+(-2)+3=0 \Rightarrow \alpha=-1$.
   4. $\bar k$-component: $2+3+(-\gamma)=0 \Rightarrow \gamma=5$.
   5. So $\overline{AB}=\bar i-\bar j+2\bar k=(1,-1,2)$ and $\overline{BC}=-3\bar i-2\bar j+3\bar k=(-3,-2,3)$.
   6. $\overline{AB}\times\overline{BC}=\left|\begin{matrix}\bar i&\bar j&\bar k\\1&-1&2\\-3&-2&3\end{matrix}\right| = \bar i[(-1)(3)-(2)(-2)]-\bar j[(1)(3)-(2)(-3)]+\bar k[(1)(-2)-(-1)(-3)]$ $=\bar i(-3+4)-\bar j(3+6)+\bar k(-2-3)=(1,-9,-5)$.
   7. $|\overline{AB}\times\overline{BC}|=\sqrt{1+81+25}=\sqrt{107}$.
   8. Area $=\dfrac12\sqrt{107}$.

   Common Mistakes

   • Using the wrong sign convention for the closure relation (e.g. $\overline{AB}+\overline{BC}=\overline{AC}$ instead of $=-\overline{CA}$).
   • Cross-product determinant sign errors, especially in the middle ($\bar j$) cofactor.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{2}\sqrt{107}$.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $A(0,1,-2), B(-1,2,-3), C(2,-3,4)$ and $D(3,4,5)$ are the vertices of a tetrahedron $ABCD$, then the volume of that tetrahedron is (A) $\dfrac{16}{3}$ (B) 32 (C) $\dfrac{8}{3}$ (D) 16

   ›Reveal solutionSolution

   The volume of a tetrahedron with vertices $A,B,C,D$ is $\dfrac16$ the absolute value of the scalar triple product of three edge vectors from one vertex. Answer: $\dfrac{16}{3}$.

   Concept and Intuition

   Three edge vectors from a common vertex of a tetrahedron span a parallelepiped whose volume is $|\overline{AB}\cdot(\overline{AC}\times\overline{AD})|$; the tetrahedron is exactly $\dfrac16$ of that parallelepiped (a standard result from decomposing the parallelepiped into 6 congruent tetrahedra).

   Step-by-Step Solution

   1. Compute edge vectors from $A(0,1,-2)$: $\overline{AB}=B-A=(-1-0,\,2-1,\,-3-(-2))=(-1,1,-1)$.
   2. $\overline{AC}=C-A=(2-0,\,-3-1,\,4-(-2))=(2,-4,6)$.
   3. $\overline{AD}=D-A=(3-0,\,4-1,\,5-(-2))=(3,3,7)$.
   4. Compute $\overline{AC}\times\overline{AD}=\begin{vmatrix}\bar i&\bar j&\bar k\\2&-4&6\\3&3&7\end{vmatrix}$: $\bar i$-comp $=(-4)(7)-(6)(3)=-28-18=-46$; $\bar j$-comp $=-[(2)(7)-(6)(3)]=-[14-18]=4$; $\bar k$-comp $=(2)(3)-(-4)(3)=6+12=18$. So $\overline{AC}\times\overline{AD}=(-46,4,18)$.
   5. Scalar triple product: $\overline{AB}\cdot(-46,4,18)=(-1)(-46)+(1)(4)+(-1)(18)=46+4-18=32$.
   6. Volume $=\dfrac16|32|=\dfrac{32}{6}=\dfrac{16}{3}$.

   Common Mistakes

   • Forgetting the factor of $\dfrac16$ and reporting the parallelepiped volume (which would give $32$, matching a distractor option pattern like doubling to $32$).
   • Sign slip in the cofactor expansion of the $\bar j$-component of the cross product.

   ✓Final answer

   The correct option is (A) — $\dfrac{16}{3}$.

   ANSWER: A

4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $|\bar a| = 2k$, $|\bar b| = k$ and $|\bar a - \bar b|^2 = 20k^2 - |2\bar a + \bar b|^2$ then, $|\bar a \times \bar b| =$ (A) $\sqrt3\, k^2$ (B) $k^2$ (C) $\dfrac{k^2}{4}$ (D) $2k$

   ›Reveal solutionSolution

   Expand both squared-magnitude expressions using $|\bar u\pm\bar v|^2=|\bar u|^2+|\bar v|^2\pm2\bar u\cdot\bar v$ to pin down $\bar a\cdot\bar b$, then find $\theta$ and use $|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta$; the answer is $\sqrt3\,k^2$.

   Concept and Intuition

   The dot product encodes the angle between two vectors ($\bar a\cdot\bar b=|\bar a||\bar b|\cos\theta$), while the cross product's magnitude encodes the same angle via sine ($|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta$). Any equation relating squared magnitudes of sums/differences of $\bar a,\bar b$ can be expanded purely in terms of $|\bar a|,|\bar b|,\bar a\cdot\bar b$ — so it's really an equation for the unknown angle in disguise.

   Step-by-Step Solution

   1. $|\bar a-\bar b|^2=|\bar a|^2+|\bar b|^2-2\bar a\cdot\bar b=(2k)^2+k^2-2\bar a\cdot\bar b=5k^2-2\bar a\cdot\bar b$.
   2. $|2\bar a+\bar b|^2=4|\bar a|^2+|\bar b|^2+4\bar a\cdot\bar b=4(4k^2)+k^2+4\bar a\cdot\bar b=17k^2+4\bar a\cdot\bar b$.
   3. The given equation is $|\bar a-\bar b|^2=20k^2-|2\bar a+\bar b|^2$, i.e. $5k^2-2\bar a\cdot\bar b=20k^2-\left(17k^2+4\bar a\cdot\bar b\right)=3k^2-4\bar a\cdot\bar b$.
   4. So $5k^2-2\bar a\cdot\bar b=3k^2-4\bar a\cdot\bar b \Rightarrow 2k^2=-2\bar a\cdot\bar b \Rightarrow \bar a\cdot\bar b=-k^2$.
   5. $\cos\theta=\dfrac{\bar a\cdot\bar b}{|\bar a||\bar b|}=\dfrac{-k^2}{(2k)(k)}=-\dfrac12$, so $\theta=120°$, $\sin\theta=\dfrac{\sqrt3}{2}$.
   6. $|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta=(2k)(k)\left(\dfrac{\sqrt3}{2}\right)=\sqrt3\,k^2$.

   Common Mistakes

   • Sign errors expanding $|2\bar a+\bar b|^2$ (the cross term is $+4\bar a\cdot\bar b$, not $-$).
   • Forgetting $\theta=120°$ gives a positive $\sin\theta$ (both $120°$ and $-120°$ have the same sine, so no ambiguity there, but some students wrongly take $\cos^{-1}(-1/2)=60°$).

   ✓Final answer

   The correct option is (A) $\sqrt3\,k^2$.

   ANSWER: A

5. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.$A(1, 2, 3)$, $B(3, 4, k)$, $C(2, 1, 4)$ form an isosceles triangle. If $AB = BC$, then the area of $\triangle ABC$ is (A) $\dfrac{\sqrt{165}}{4}$ (B) $\dfrac{15}{4}$ (C) $\dfrac{7}{2}$ (D) $\dfrac{\sqrt{114}}{4}$

   ›Reveal solutionSolution

   This tests solving for an unknown coordinate using the isosceles condition $AB=BC$, then computing the area of a 3D triangle via the cross product of two side vectors. The area comes out to $\dfrac{\sqrt{114}}4$.

   Concept and Intuition

   For points in 3D, distances are found the same way as in 2D but with an extra coordinate; setting two side-lengths equal (here $AB=BC$) gives a single equation in the unknown $k$. Once all three vertices are fully known, the area of the triangle is most efficiently computed as $\dfrac12\left|\vec{AB}\times\vec{AC}\right|$ — the cross product's magnitude gives twice the triangle's area regardless of orientation in 3D space (unlike the 2D shoelace formula, which needs the points to be coplanar with the $xy$-plane).

   Step-by-Step Solution

   1. $A(1,2,3),\,B(3,4,k),\,C(2,1,4)$.
   2. $AB^2=(3-1)^2+(4-2)^2+(k-3)^2=4+4+(k-3)^2=8+(k-3)^2$.
   3. $BC^2=(2-3)^2+(1-4)^2+(4-k)^2=1+9+(k-4)^2=10+(k-4)^2$.
   4. Set $AB^2=BC^2$:

   $8+(k-3)^2=10+(k-4)^2$

   $8+k^2-6k+9=10+k^2-8k+16$

   $17-6k=26-8k$

   $2k=9\;\Rightarrow\;k=\frac92.$

   5. So $B=\left(3,4,\dfrac92\right)$.
   6. $\vec{AB}=B-A=(2,\,2,\,1.5)$, $\vec{AC}=C-A=(1,\,-1,\,1)$.
   7. Cross product $\vec{AB}\times\vec{AC}$:

   $\left(AB_yAC_z-AB_zAC_y,\;AB_zAC_x-AB_xAC_z,\;AB_xAC_y-AB_yAC_x\right)$

   $=(2(1)-1.5(-1),\;1.5(1)-2(1),\;2(-1)-2(1))=(2+1.5,\;1.5-2,\;-2-2)=(3.5,\,-0.5,\,-4).$

   8. Magnitude: $\sqrt{3.5^2+0.5^2+4^2}=\sqrt{12.25+0.25+16}=\sqrt{28.5}=\sqrt{\dfrac{57}2}=\dfrac{\sqrt{114}}2$.
   9. Area $=\dfrac12\left|\vec{AB}\times\vec{AC}\right|=\dfrac12\cdot\dfrac{\sqrt{114}}2=\dfrac{\sqrt{114}}4$.

   Common Mistakes

   • Setting up $AB=BC$ incorrectly (e.g. accidentally using $AC=BC$ instead, which is a different, unintended condition).
   • Arithmetic slips solving the linear equation for $k$ (easy to mis-expand $(k-3)^2$ vs $(k-4)^2$).
   • Cross-product component sign errors — always use the cyclic pattern $(y_1z_2-z_1y_2,\ z_1x_2-x_1z_2,\ x_1y_2-y_1x_2)$ consistently.

   ✓Final answer

   The correct option is (D) — $\dfrac{\sqrt{114}}{4}$.

   ANSWER: D

6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If $\bar{a} = \bar{i} + p\bar{j} - 3\bar{k}$, $\bar{b} = p\bar{i} - 3\bar{j} + \bar{k}$, $\bar{c} = -3\bar{i} + \bar{j} + 2\bar{k}$ are three vectors such that $|\bar{a}\times\bar{b}| = |\bar{a}\times\bar{c}|$, then $p =$ (A) $-2$ (B) $-1$ (C) $1$ (D) $2$

   ›Reveal solutionSolution

   This tests the cross product and vector magnitude — set up $\bar a\times\bar b$ and $\bar a\times\bar c$ componentwise and solve $|\bar a\times\bar b|=|\bar a\times\bar c|$ for $p$. The answer is $p=2$.

   Concept and Intuition

   $|\bar a\times\bar b|$ and $|\bar a\times\bar c|$ are the areas of the parallelograms spanned by each pair of vectors. Setting the two areas equal gives one equation in the unknown $p$. The cleanest route is to compute both cross products in components, square their magnitudes, and simplify — any common term on both sides cancels immediately, which is exactly what happens here.

   Step-by-Step Solution

   1. Vectors: $\bar a=(1,p,-3)$, $\bar b=(p,-3,1)$, $\bar c=(-3,1,2)$.
   2. Compute $\bar a\times\bar b$ using $(a_yb_z-a_zb_y,\ a_zb_x-a_xb_z,\ a_xb_y-a_yb_x)$:

   $\bar a\times\bar b=(p(1)-(-3)(-3),\ (-3)(p)-(1)(1),\ (1)(-3)-(p)(p))=(p-9,\,-3p-1,\,-3-p^2).$

   3. Compute $\bar a\times\bar c$:

   $\bar a\times\bar c=(p(2)-(-3)(1),\ (-3)(-3)-(1)(2),\ (1)(1)-(p)(-3))=(2p+3,\,7,\,1+3p).$

   4. Set $|\bar a\times\bar b|^2=|\bar a\times\bar c|^2$:

   $(p-9)^2+(3p+1)^2+(3+p^2)^2=(2p+3)^2+7^2+(1+3p)^2.$

   5. Note $(3p+1)^2=(1+3p)^2$ appears on both sides, so it cancels:

   $(p-9)^2+(3+p^2)^2=(2p+3)^2+49.$

   6. Expand: $(p-9)^2=p^2-18p+81$; $(3+p^2)^2=p^4+6p^2+9$. Sum $=p^4+7p^2-18p+90$. RHS: $(2p+3)^2+49=4p^2+12p+9+49=4p^2+12p+58$.
   7. So $p^4+7p^2-18p+90=4p^2+12p+58 \Rightarrow p^4+3p^2-30p+32=0$.
   8. Test $p=2$: $16+12-60+32=0$ ✓. (Checking the others: $p=-2\to120$, $p=-1\to66$, $p=1\to6$ — none vanish.) So $p=2$.

   Common Mistakes

   • Sign errors while expanding the cross-product components (easy to flip a sign in the $y$-component).
   • Forgetting that the middle term cancels, leading to a needlessly messy quartic.
   • Trying to factor the quartic algebraically instead of testing the four given roots — with MCQ options given, direct substitution is faster and safer.

   ✓Final answer

   The correct option is (D) — $p=2$.

   ANSWER: D

7. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\bar{a}=2\bar{i}-\bar{j}+6\bar{k}$; $\bar{b}=\bar{i}-\bar{j}+\bar{k}$ and $\bar{c}=3\bar{j}-\bar{k}$, then $\bar{a}\times\bar{b}+\bar{b}\times\bar{c}+\bar{c}\times\bar{a} =$ (A) $20\bar{i}+3\bar{j}-4\bar{k}$ (B) $20\bar{i}-3\bar{j}+4\bar{k}$ (C) $3\bar{i}+20\bar{j}-4\bar{k}$ (D) $4\bar{i}+20\bar{j}-3\bar{k}$

   ›Reveal solutionSolution

   Direct componentwise computation of the three cross products and their sum gives $20\bar i+3\bar j-4\bar k$.

   Concept and Intuition

   There's no shortcut identity needed here beyond careful, direct computation of each $2\times2$ minor for the three cross products, then adding the resulting vectors componentwise.

   Step-by-Step Solution

   1. $\bar a=(2,-1,6),\ \bar b=(1,-1,1),\ \bar c=(0,3,-1)$.
   2. $\bar a\times\bar b = \big((-1)(1)-(6)(-1),\ -[(2)(1)-(6)(1)],\ (2)(-1)-(-1)(1)\big) = (5,\ 4,\ -1)$.
   3. $\bar b\times\bar c = \big((-1)(-1)-(1)(3),\ -[(1)(-1)-(1)(0)],\ (1)(3)-(-1)(0)\big) = (-2,\ 1,\ 3)$.
   4. $\bar c\times\bar a = \big((3)(6)-(-1)(-1),\ -[(0)(6)-(-1)(2)],\ (0)(-1)-(3)(2)\big) = (17,\ -2,\ -6)$.
   5. Sum: $(5-2+17,\ 4+1-2,\ -1+3-6) = (20,\ 3,\ -4)$.

   Common Mistakes

   • Sign errors in the $j$-component of a cross product (it carries a negative sign in the cofactor expansion) — the most common slip in these problems.
   • Mixing up the order in $\bar b\times\bar c$ vs $\bar c\times\bar b$, which flips the sign of that term.

   ✓Final answer

   The correct option is (A) — $20\bar i+3\bar j-4\bar k$.

   ANSWER: A

8. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If the position vectors of the vertices A, B, C of a triangle are $3\bar{i}+4\bar{j}-\bar{k}$, $\bar{i}+3\bar{j}+\bar{k}$, $5(\bar{i}+\bar{j}+\bar{k})$ respectively, then the magnitude of the altitude drawn from A on to the side BC is (A) $\dfrac{4\sqrt{5}}{3}$ (B) $\dfrac{5\sqrt{5}}{3}$ (C) $\dfrac{7\sqrt{5}}{3}$ (D) $\dfrac{8\sqrt{5}}{3}$

   ›Reveal solutionSolution

   Computing the triangle's area via the cross product $\vec{BA}\times\vec{BC}$ and dividing by the base $|BC|$ gives the altitude from $A$ as $\dfrac{4\sqrt5}3$.

   Concept and Intuition

   The perpendicular distance (altitude) from a vertex to the opposite side of a triangle is most directly found via $\text{Area}=\tfrac12\times\text{base}\times\text{height}$, where the area itself comes from $\tfrac12|\vec u\times\vec v|$ for any two vectors along two sides from a common vertex.

   Step-by-Step Solution

   1. Given $A=3i+4j-k,\ B=i+3j+k,\ C=5i+5j+5k$.
   2. $\vec{BC}=C-B=(5-1)i+(5-3)j+(5-1)k=4i+2j+4k$; $|\vec{BC}|=\sqrt{16+4+16}=\sqrt{36}=6$.
   3. $\vec{BA}=A-B=(3-1)i+(4-3)j+(-1-1)k=2i+j-2k$.
   4. Cross product $\vec{BA}\times\vec{BC}=\begin{vmatrix}i&j&k\\2&1&-2\\4&2&4\end{vmatrix}$:
      • $i$-component: $1\cdot4-(-2)\cdot2=4+4=8$
      • $j$-component: $-(2\cdot4-(-2)\cdot4)=-(8+8)=-16$
      • $k$-component: $2\cdot2-1\cdot4=4-4=0$ So $\vec{BA}\times\vec{BC}=(8,-16,0)$, with magnitude $\sqrt{64+256+0}=\sqrt{320}=8\sqrt5$.
   5. Area of $\triangle ABC=\tfrac12\times8\sqrt5=4\sqrt5$.
   6. Using $BC$ as base: $\text{Area}=\tfrac12\cdot|BC|\cdot h\Rightarrow4\sqrt5=\tfrac12\cdot6\cdot h=3h\Rightarrow h=\dfrac{4\sqrt5}3$.

   Common Mistakes

   • Computing the cross product with a sign error in the $j$-component (forgetting the minus sign in the cofactor expansion).
   • Using $|AB|$ or $|AC|$ instead of $|BC|$ as the base when the altitude asked for is from $A$ onto $BC$.

   ✓Final answer

   The correct option is (A) — $\dfrac{4\sqrt5}3$.

   ANSWER: A

9. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\bar a$, $\bar b$ and $\bar c$ are the position vectors of three non-collinear points on a plane. If $\alpha = [\bar a\ \ \bar b\ \ \bar c]$ and $\bar r = \bar a \times \bar b - \bar c \times \bar b - \bar a \times \bar c$, then $\dfrac{|\alpha|}{|\bar r|}$ represents (A) Ratio of areas of the triangles formed by $\bar o, \bar a, \bar b$ to $\bar o, \bar b, \bar c$ (B) Ratio of the numerical values of volume of the parallelopiped formed with $\bar o, \bar a, \bar b, \bar c$ and its height (C) Ratio of lengths of the diagonals of the parallelopiped formed with $\bar o, \bar a, \bar b, \bar c$ (D) Length of the perpendicular from origin to the plane

   ›Reveal solutionSolution

   $\bar r$ turns out to be twice the (origin-independent) area vector of $\triangle ABC$, while $\alpha$ is $6\times$ the volume of tetrahedron $OABC$; their ratio collapses to exactly the perpendicular distance from $O$ to the plane $ABC$.

   Concept and Intuition

   For position vectors $\bar a,\bar b,\bar c$ of a triangle's vertices (measured from any origin $O$), the combination $\bar a\times\bar b+\bar b\times\bar c+\bar c\times\bar a$ is a fixed vector normal to the plane $ABC$ whose magnitude is $2\times$Area($ABC$) — this is independent of where $O$ is, because it is really just the sum of oriented areas of triangles $OAB$, $OBC$, $OCA$, which telescopes into the area of $ABC$ itself. Meanwhile the scalar triple product $[\bar a\ \bar b\ \bar c]$ measures $6\times$ the volume of tetrahedron $OABC$. Comparing a volume-based quantity to an area-based quantity naturally produces a length — the height of that tetrahedron from $O$.

   Step-by-Step Solution

   1. Rewrite $\bar r$: $-\bar c\times\bar b = \bar b\times\bar c$ and $-\bar a\times\bar c=\bar c\times\bar a$, so $\bar r = \bar a\times\bar b+\bar b\times\bar c+\bar c\times\bar a$.
   2. This is the standard "twice area vector" formula for triangle $ABC$: $|\bar r| = 2\,\text{Area}(ABC)$ (true regardless of the choice of origin $O$).
   3. $\alpha = [\bar a\ \bar b\ \bar c] = \bar a\cdot(\bar b\times\bar c)$. The volume of tetrahedron $OABC$ is $V=\dfrac16|\alpha|$, so $|\alpha| = 6V$.
   4. Also, treating $\triangle ABC$ as the base of the tetrahedron with apex $O$: $V = \dfrac13\,\text{Area}(ABC)\cdot h$, where $h$ is the perpendicular distance from $O$ to plane $ABC$. So $3V = \text{Area}(ABC)\cdot h$, giving $|\alpha| = 6V = 2\,\text{Area}(ABC)\cdot h$.
   5. Therefore $\dfrac{|\alpha|}{|\bar r|} = \dfrac{2\,\text{Area}(ABC)\cdot h}{2\,\text{Area}(ABC)} = h$.
   6. $h$ is exactly the perpendicular distance from the origin to the plane $ABC$.

   Common Mistakes

   • Missing the sign flips ($-\bar c\times\bar b=\bar b\times\bar c$) and failing to recognise the standard "sum of cross products" area formula.
   • Confusing the tetrahedron volume formula's factor of $\frac13$ vs $\frac16$.

   ✓Final answer

   The correct option is (D) — Length of the perpendicular from origin to the plane.

   ANSWER: D

10. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.If $\bar{a}$ and $\bar{b}$ are two vectors such that $|\bar{a}| = |\bar{b}| = \sqrt{14}$ and $\bar{a}.\bar{b} = -7$, then $\dfrac{|\bar{a} \times \bar{b}|}{|\bar{a}.\bar{b}|} =$ (A) $7\sqrt{3}$ (B) $\sqrt{3}$ (C) $49\sqrt{3}$ (D) $\dfrac{\sqrt{3}}{7}$

    ›Reveal solutionSolution

    Using $\bar a\cdot\bar b=|\bar a||\bar b|\cos\theta$ to find $\theta$, then $|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta$, the ratio simplifies to $\sqrt3$.

    Concept and Intuition

    The dot product encodes $\cos\theta$ between two vectors, and the cross product's magnitude encodes $\sin\theta$ (scaled by the same product of magnitudes). Dividing one by the other cancels the magnitudes and leaves a pure trig ratio, once we know $\theta$.

    Step-by-Step Solution

    1. $|\bar a|=|\bar b|=\sqrt{14}$, so $|\bar a||\bar b|=14$.
    2. $\cos\theta=\dfrac{\bar a\cdot\bar b}{|\bar a||\bar b|}=\dfrac{-7}{14}=-\dfrac12\Rightarrow\theta=120^\circ$.
    3. $\sin120^\circ=\dfrac{\sqrt3}{2}$.
    4. $|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta=14\times\dfrac{\sqrt3}{2}=7\sqrt3$.
    5. $|\bar a\cdot\bar b|=7$.
    6. Ratio $=\dfrac{7\sqrt3}{7}=\sqrt3$.

    Common Mistakes

    • Taking $\theta=60^\circ$ instead of $120^\circ$ by dropping the negative sign of $\cos\theta$ — the angle between vectors can be obtuse, and $\sin$ is still positive either way but it's important to get $\cos\theta$ right if it were ever used further.

    ✓Final answer

    The correct option is (B) — $\sqrt{3}$.

    ANSWER: B

11. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.Let $\bar{a} = \bar{i}+2\bar{j}+3\bar{k}$ and $\bar{b} = \bar{i}-2\bar{j}-3\bar{k}$ be two vectors. If $A_1$ is the area of the quadrilateral having $\bar{a}, \bar{b}$ as its diagonals and $A_2$ is the area of the parallelogram having $\bar{a}, \bar{b}$ as its two adjacent sides, then $A_1.A_2 =$ (A) $26$ (B) $\dfrac{27}{2}$ (C) $52$ (D) $27$

    ›Reveal solutionSolution

    Both areas reduce to expressions in $|\bar a\times\bar b|$; computing the cross product and combining gives $A_1A_2=26$.

    Concept and Intuition

    For a quadrilateral whose diagonals are given by vectors $\bar a,\bar b$, its area is $A_1=\tfrac12|\bar a\times\bar b|$ (a standard vector-geometry identity, since the diagonals split the quadrilateral into four triangles whose combined area works out to half the diagonal cross-product magnitude). For a parallelogram with adjacent sides $\bar a,\bar b$, the area is $A_2=|\bar a\times\bar b|$. So the product $A_1A_2$ is just $\tfrac12|\bar a\times\bar b|^2$ — everything reduces to one cross-product computation.

    Step-by-Step Solution

    1. $\bar a=(1,2,3)$, $\bar b=(1,-2,-3)$.
    2. $\bar a\times\bar b = \begin{vmatrix}\bar i&\bar j&\bar k\\1&2&3\\1&-2&-3\end{vmatrix} = \bar i[(2)(-3)-(3)(-2)] - \bar j[(1)(-3)-(3)(1)] + \bar k[(1)(-2)-(2)(1)]$ $= \bar i[-6+6] - \bar j[-3-3] + \bar k[-2-2] = (0,6,-4)$.
    3. $|\bar a\times\bar b|^2 = 0^2+6^2+(-4)^2 = 36+16=52$, so $|\bar a\times\bar b|=\sqrt{52}=2\sqrt{13}$.
    4. $A_1 = \tfrac12(2\sqrt{13}) = \sqrt{13}$; $A_2 = 2\sqrt{13}$.
    5. $A_1 A_2 = \sqrt{13}\times 2\sqrt{13} = 2\times13 = 26$.

    Common Mistakes

    • Using $A_1=|\bar a\times\bar b|$ (forgetting the factor of $\tfrac12$ for the diagonal-quadrilateral formula) — that formula's factor of $\tfrac12$ is specific to using the diagonals, distinct from the side-based parallelogram formula.
    • Arithmetic slips in the $3\times3$ cross-product expansion.

    ✓Final answer

    The correct option is (A) — $26$.

    ANSWER: A

12. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.For some real number $\lambda$, if the area of the triangle having $\bar{a} = 3\bar{i}-\bar{j}+\lambda\bar{k}$ and $\bar{b} = \lambda\bar{i}+\bar{j}-3\bar{k}$ as two of its sides is $\dfrac{\sqrt{195}}{2}$, then the number of distinct possible values of $\lambda$ is (A) $4$ (B) $3$ (C) $2$ (D) $1$

    ›Reveal solutionSolution

    Setting $|\bar a\times\bar b|^2=195$ leads to a quadratic in $\lambda^2$ with only one valid (non-negative) root, giving exactly two real values of $\lambda$.

    Concept and Intuition

    The area of a triangle with two sides given by vectors $\bar a,\bar b$ (from a common vertex) is $\tfrac12|\bar a\times\bar b|$. Setting this equal to the given area produces an equation in $\lambda$ through the magnitude of the cross product. Since the cross-product components involve $\lambda^2$ symmetrically, it's natural to substitute $u=\lambda^2$ to reduce the resulting quartic to a quadratic.

    Step-by-Step Solution

    1. $\bar a=(3,-1,\lambda)$, $\bar b=(\lambda,1,-3)$. Compute the cross product: $\bar a\times\bar b = \big((-1)(-3)-(\lambda)(1),\ -[(3)(-3)-(\lambda)(\lambda)],\ (3)(1)-(-1)(\lambda)\big) = (3-\lambda,\ 9+\lambda^2,\ 3+\lambda)$.
    2. Area condition: $\tfrac12|\bar a\times\bar b| = \dfrac{\sqrt{195}}2 \Rightarrow |\bar a\times\bar b|^2 = 195$.
    3. $(3-\lambda)^2+(3+\lambda)^2 = 2(9+\lambda^2) = 18+2\lambda^2$ (sum-of-squares identity).
    4. So $18+2\lambda^2+(9+\lambda^2)^2 = 195$. Let $u=\lambda^2$: $18+2u+(9+u)^2=195 \Rightarrow 18+2u+81+18u+u^2=195 \Rightarrow u^2+20u+99=195 \Rightarrow u^2+20u-96=0$.
    5. Solve: $u=\dfrac{-20\pm\sqrt{400+384}}2=\dfrac{-20\pm28}2$, giving $u=4$ or $u=-24$.
    6. Since $u=\lambda^2\ge0$, reject $u=-24$; keep $u=4 \Rightarrow \lambda=\pm2$ — exactly 2 distinct real values.

    Common Mistakes

    • Forgetting to discard the negative root of $u=\lambda^2$ (which would otherwise wrongly suggest 4 solutions if complex roots were counted).
    • Sign errors when expanding the cofactors of the $3\times3$ determinant for the cross product.

    ✓Final answer

    The correct option is (C) — $2$.

    ANSWER: C