Q.Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat i-\hat j+3\hat k$ and $\vec{b}=2\hat i-7\hat j+\hat k$.

Showing the 12 most recent of 20 on this concept.

1. 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. $\vec u\times\vec v=(5,5,-5)$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $|\bar a\times\bar b|=\sqrt{52}=2\sqrt{13}$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $\vec a=\hat i+\hat j$ and $\vec b=\hat j+\hat k$ are two vectors, then $|\vec a \times \vec b|=$ (A) 0 (B) 3 (C) $\sqrt3$ (D) 1

   ›Reveal solutionSolution

   A direct determinant cross-product computation gives $\vec a\times\vec b=(1,-1,1)$, whose magnitude is $\sqrt3$.

   Concept and Intuition

   The cross product of two vectors given in component form is computed via the standard $3\times3$ determinant with $\hat i,\hat j,\hat k$ in the first row.

   Step-by-Step Solution

   1. $\vec a=(1,1,0)$, $\vec b=(0,1,1)$.
   2. $\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\\1&1&0\\0&1&1\end{vmatrix} = \hat i(1\cdot1-0\cdot1)-\hat j(1\cdot1-0\cdot0)+\hat k(1\cdot1-1\cdot0)$
   3. $=\hat i(1)-\hat j(1)+\hat k(1) = \hat i-\hat j+\hat k$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $\overline{BC}=-3\bar i-2\bar j+3\bar k=(-3,-2,3)$. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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. $|\bar a\cdot\bar b|=7$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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 $\sqrt{64+256+0}=\sqrt{320}=8\sqrt5$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $5k^2-2\bar a\cdot\bar b=3k^2-4\bar a\cdot\bar b \Rightarrow 2k^2=-2\bar a\cdot\bar b \Rightarrow \bar a\cdot\bar b=-k^2$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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. $\bar a\times\bar c=\begin{vmatrix}\bar i&\bar j&\bar k\\2&3&-1\\-1/2&1&-1\end{vmatrix}=\bar i(-3+1)-\bar j(-2-\tfrac12)+\bar k(2+\tfrac32)=(-2,\,\tfrac52,\,\tfrac72)$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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). …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $|\alpha| = 6V$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.Let $x, y$ are real numbers. If $\vec{a} = (\sin x)\hat{i} + (\sin y)\hat{j}$ and $\vec{b} = (\cos x)\hat{i} + (\cos y)\hat{j}$, then $|\vec{a}\times\vec{b}|$ is (A) 0 (B) Greater than one (C) Less than or equal to 1 (D) Less than 1

    ›Reveal solutionSolution

    This tests the cross product of two unit-length-style vectors built from sine/cosine components; it collapses to $|\sin(x-y)|$, which is bounded by 1.

    Concept and Intuition

    For planar vectors $\vec a=a_1\hat i+a_2\hat j$ and $\vec b=b_1\hat i+b_2\hat j$, the magnitude of the cross product is $|a_1b_2-a_2b_1|$ (the $\hat k$-component). Recognizing a trig identity inside this expression is the key move.

    Step-by-Step Solution

    1. Here $a_1=\sin x,\ a_2=\sin y,\ b_1=\cos x,\ b_2=\cos y$.
    2. $\vec a\times\vec b = (\sin x\cos y - \sin y\cos x)\hat k = \sin(x-y)\hat k$.
    3. So $|\vec a\times\vec b| = |\sin(x-y)|$.
    4. Since sine of any real angle lies in $[-1,1]$, we always have $|\sin(x-y)|\le 1$; it can be less than 1, equal to 1, or even 0 depending on $x,y$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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 $\vec{AB}\times\vec{AC}$:

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

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

    🔒 Start your 14-day free trial to see the full solution →