Q.Find the area of the triangle with vertices $A(1, 1, 2)$, $B(2, 3, 5)$ and $C(1, 5, 5)$.

Showing the 12 most recent of 20 on this concept.

1. 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 $\sqrt{64+256+0}=\sqrt{320}=8\sqrt5$. …

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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 $\overline{BC}=-3\bar i-2\bar j+3\bar k=(-3,-2,3)$. …

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3. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The area of the parallelogram for which the vectors $i+j+2k$ and $3i-2j+k$ are adjacent sides is equal to (A) $3\sqrt{5}$ (B) $5\sqrt{3}$ (C) $2\sqrt{5}$ (D) $5\sqrt{6}$

   ›Reveal solutionSolution

   The parallelogram area is the magnitude of the cross product of the two adjacent side vectors, which computes to $5\sqrt3$.

   Concept and Intuition

   For a parallelogram with adjacent sides given by vectors $\vec u$ and $\vec v$, the area equals $|\vec u\times\vec v|$ — the cross product magnitude directly measures the parallelogram's area.

   Step-by-Step Solution

   1. $\vec u=\hat i+\hat j+2\hat k=(1,1,2)$, $\vec v=3\hat i-2\hat j+\hat k=(3,-2,1)$.
   2. $\vec u\times\vec v = \begin{vmatrix}\hat i&\hat j&\hat k\\1&1&2\\3&-2&1\end{vmatrix}$.
   3. $\hat i$ component: $(1)(1)-(2)(-2)=1+4=5$.
   4. $\hat j$ component: $-[(1)(1)-(2)(3)]=-[1-6]=5$.
   5. $\hat k$ component: $(1)(-2)-(1)(3)=-2-3=-5$.
   6. $\vec u\times\vec v=(5,5,-5)$. …

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4. 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 $\overline{AC}\times\overline{AD}=(-46,4,18)$. …

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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 $\vec{AB}\times\vec{AC}$:

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

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

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6. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.A, B, C, D are any 4 points and $\left|\overline{AB}\times\overline{CD} + \overline{BC}\times\overline{AD} + \overline{CA}\times\overline{BD}\right| = \lambda (\text{Area of } \triangle ABC)$ then $\lambda =$ (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   Tests a classic 4-point cross-product identity that reduces to 4 times the triangle's area; answer is $\lambda=4$.

   Concept and Intuition

   Each term like $\overline{AB}\times\overline{CD}$ can be rewritten in terms of position vectors $a,b,c,d$. Expanding all three cyclic terms causes most cross-products to cancel in pairs, leaving an expression proportional to $\overline{AB}\times\overline{AC}$ — exactly the vector whose magnitude gives twice the area of triangle $ABC$.

   Step-by-Step Solution

   1. Let position vectors be $a,b,c,d$ for $A,B,C,D$. Write: $\overline{AB}\times\overline{CD}=(b-a)\times(d-c)$, and similarly for the other two terms.
   2. Expanding and adding all three cyclic terms, the $b\times d$, $a\times d$, and $c\times d$ contributions cancel pairwise, leaving $\text{Sum} = 2(a\times c - a\times b - b\times c)$.
   3. Separately, $\overline{AB}\times\overline{AC}=(b-a)\times(c-a)=b\times c+a\times b-a\times c = -(a\times c-a\times b-b\times c)$.
   4. So $\text{Sum} = -2(\overline{AB}\times\overline{AC})$, hence $|\text{Sum}|=2|\overline{AB}\times\overline{AC}|$. …

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7. 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 $|\bar a\times\bar b|=\sqrt{52}=2\sqrt{13}$. …

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8. 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 $|\alpha| = 6V$. …

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9. 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. $100+(2d+12)^2+(d-9)^2=600$. …

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10. 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). …

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11. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.If the vectors $\bar a=2\bar i+3\bar j-\bar k$, $\bar b=4\bar i-\bar j+3\bar k$ and $\bar c=p\bar i+\bar j-\bar k$ are coplanar, then $|\bar a\times\bar c|=$ (A) $\sqrt{14}$ (B) $\dfrac{3\sqrt{10}}{2}$ (C) $\sqrt{26}$ (D) $\dfrac{\sqrt{90}}{4}$

    ›Reveal solutionSolution

    Coplanarity fixes $p=-\tfrac12$ via the scalar triple product; then $|\bar a\times\bar c|=\dfrac{3\sqrt{10}}{2}$.

    Concept and Intuition

    Three vectors are coplanar exactly when their scalar triple product vanishes: $\bar a\cdot(\bar b\times\bar c)=0$. This gives one equation to solve for the unknown $p$ in $\bar c$, after which the required cross product is a direct computation.

    Step-by-Step Solution

    1. $\bar b\times\bar c=\begin{vmatrix}\bar i&\bar j&\bar k\\4&-1&3\\p&1&-1\end{vmatrix}=\bar i(1-3)-\bar j(-4-3p)+\bar k(4+p)=(-2,\,4+3p,\,4+p)$.
    2. $\bar a\cdot(\bar b\times\bar c)=2(-2)+3(4+3p)+(-1)(4+p)=-4+12+9p-4-p=4+8p$.
    3. Set to $0$: $4+8p=0\Rightarrow p=-\dfrac12$, so $\bar c=-\tfrac12\bar i+\bar j-\bar k$.
    4. $\bar a\times\bar c=\begin{vmatrix}\bar i&\bar j&\bar k\\2&3&-1\\-1/2&1&-1\end{vmatrix}=\bar i(-3+1)-\bar j(-2-\tfrac12)+\bar k(2+\tfrac32)=(-2,\,\tfrac52,\,\tfrac72)$. …

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12. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.If $|a| = 13$, $|b| = 5$ and $\bar{a}.\bar{b} = 60$ then $|\bar{a} \times \bar{b}| =$ (A) 15 (B) 20 (C) 30 (D) 25

    ›Reveal solutionSolution

    The Lagrange identity connects dot and cross products directly — the answer is (D) 25.

    Concept and Intuition

    For any two vectors, $|\bar a\times\bar b|^2+(\bar a\cdot\bar b)^2 = |\bar a|^2|\bar b|^2$ (this follows from $|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta$ and $\bar a\cdot\bar b=|\bar a||\bar b|\cos\theta$, using $\sin^2\theta+\cos^2\theta=1$).

    Step-by-Step Solution

    1. Use the identity: $|\bar a\times\bar b|^2 = |\bar a|^2|\bar b|^2 - (\bar a\cdot\bar b)^2$. …

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