Q.Three vertices of a Parallelogram ABCD are $A(1,2,3)$, $B(-1,-2,-1)$ and $C(2,3,2)$. Find the fourth vertex D.

Showing the 12 most recent of 44 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If the equation of the plane making equal intercepts on the coordinate axes and which is at a distance of $5\sqrt{3}$ units from origin is $x + y + z = k$ $(k > 0)$, then $k =$ (A) $15$ (B) $5$ (C) $\dfrac{5}{\sqrt{3}}$ (D) $15\sqrt{3}$

   ›Reveal solutionSolution

   Using the point-to-plane distance formula on $x+y+z=k$ directly gives $k=15$.

   Concept and Intuition

   A plane making equal intercepts $k$ on all three axes is $\dfrac{x}{k}+\dfrac{y}{k}+\dfrac{z}{k}=1$, i.e. $x+y+z=k$. The perpendicular distance from the origin to $Ax+By+Cz=D$ is $\dfrac{|D|}{\sqrt{A^2+B^2+C^2}}$.

   Step-by-Step Solution

   1. Plane: $x+y+z=k$, so $A=B=C=1$, $D=k$.
   2. Distance from origin $=\dfrac{|k|}{\sqrt{1+1+1}}=\dfrac{k}{\sqrt3}$ (taking $k>0$ as given). …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $A(1,1,1), B(2,3,4)$ and $C(2,5,7)$ are the vertices of $\triangle ABC$, then the length of the altitude drawn through the vertex $A$ is (A) $2$ (B) $1$ (C) $\sqrt{2}$ (D) $\sqrt{3}$

   ›Reveal solutionSolution

   Using the cross-product to get the triangle's area and dividing by $|BC|$ gives an altitude of exactly $1$. Answer: (B).

   Concept and Intuition

   The altitude from a vertex to the opposite side is simply twice the triangle's area divided by the length of that side — a direct consequence of the area formula $\text{Area}=\tfrac12\times\text{base}\times\text{height}$. In 3D, the area is most efficiently computed via the cross product of two side vectors.

   Step-by-Step Solution

   1. $\vec{AB}=B-A=(1,2,3)$, $\vec{AC}=C-A=(1,4,6)$.
   2. $\vec{AB}\times\vec{AC}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&3\\1&4&6\end{vmatrix}=\mathbf{i}(12-12)-\mathbf{j}(6-3)+\mathbf{k}(4-2)=(0,-3,2)$
   3. $|\vec{AB}\times\vec{AC}|=\sqrt{0+9+4}=\sqrt{13}$, so Area $=\dfrac12\sqrt{13}$.
   4. $\vec{BC}=C-B=(0,2,3)$, $|BC|=\sqrt{0+4+9}=\sqrt{13}$. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.$A(3,1,2)$, $B(-1,6,2)$ and $C(1,1,-2)$ are three points. A plane is passing through A and is perpendicular to the line joining B and C. If $(\alpha,\beta,\gamma)$ is the image of C with respect to the plane, then $3\alpha-\gamma+2\beta=$ (A) $0$ (B) $6$ (C) $7$ (D) $5$

   ›Reveal solutionSolution

   The plane's normal is along $BC$; reflecting $C$ in that plane and combining coordinates as asked gives $3\alpha-\gamma+2\beta=7$.

   Concept and Intuition

   "A plane perpendicular to line $BC$" means the direction vector of $BC$ is the plane's normal vector. Once we have the plane equation, the reflection (image) of any point in it is found by moving the point along the normal direction by twice its signed distance to the plane.

   Step-by-Step Solution

   1. Normal direction: $\vec n = C-B = (1-(-1),\,1-6,\,-2-2)=(2,-5,-4)$.
   2. Plane through $A(3,1,2)$ with this normal: $2(x-3)-5(y-1)-4(z-2)=0$.
   3. Expand: $2x-6-5y+5-4z+8=0\Rightarrow 2x-5y-4z+7=0$. (So $d=7$.)
   4. Reflect $C(1,1,-2)$: compute $n\cdot C + d = 2(1)-5(1)-4(-2)+7 = 2-5+8+7=12$.
   5. $|n|^2 = 4+25+16=45$.
   6. Scale factor $t = \dfrac{2(12)}{45}=\dfrac{24}{45}=\dfrac{8}{15}$.
   7. Image $= C - t\,n = (1,1,-2) - \dfrac{8}{15}(2,-5,-4)$.
   8. $x$: $1-\dfrac{16}{15}=-\dfrac1{15}$; $y$: $1+\dfrac{40}{15}=\dfrac{55}{15}=\dfrac{11}{3}$; $z$: $-2+\dfrac{32}{15}=\dfrac{2}{15}$. …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If the feet of the perpendiculars drawn from the point $(3, 4, 5)$ to the X, Y, Z – coordinate axes are $A, B, C$ respectively and the angle between $AB$ and $AC$ is $\mathrm{Cos}^{-1}\left(\frac{9}{a}\right)$, then $a =$ (A) $5\sqrt{34}$ (B) $3\sqrt{34}$ (C) $2\sqrt{34}$ (D) $\sqrt{34}$

   ›Reveal solutionSolution

   This tests finding the angle between two vectors formed from projections onto coordinate axes. Computing the dot product and magnitudes gives $a=5\sqrt{34}$.

   Concept and Intuition

   The foot of the perpendicular from a point to a coordinate axis simply keeps that axis's coordinate and zeroes the other two. Once $A,B,C$ are known, the angle at $A$ in triangle $ABC$ is found by the standard dot-product formula applied to $\vec{AB}$ and $\vec{AC}$.

   Step-by-Step Solution

   1. Foot on X-axis: $A=(3,0,0)$. Foot on Y-axis: $B=(0,4,0)$. Foot on Z-axis: $C=(0,0,5)$.
   2. $\vec{AB}=B-A=(-3,4,0)$, $\vec{AC}=C-A=(-3,0,5)$.
   3. $\vec{AB}\cdot\vec{AC}=(-3)(-3)+4(0)+0(5)=9$.
   4. $|\vec{AB}|=\sqrt{9+16+0}=5$, $|\vec{AC}|=\sqrt{9+0+25}=\sqrt{34}$.
   5. $\cos\theta=\frac{9}{5\sqrt{34}}$. …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If the equation of the plane containing the line $\bar{r} = \bar{i}+2\bar{j}+\bar{k}+t(\bar{i}-\bar{j}+2\bar{k})$ and parallel to the line $\bar{r} = -\bar{i}+2\bar{j}+s(-\bar{i}+2\bar{j}+\bar{k})$ in Cartesian coordinates is $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$, then $a+3b+c =$ (A) $2$ (B) $10$ (C) $-5$ (D) $12$

   ›Reveal solutionSolution

   Build the plane from a point + two directions (the line's own direction and the parallel line's direction), then read off intercepts.

   Concept and Intuition

   A plane containing a given line and parallel to another given line is fully determined: its normal must be perpendicular to both direction vectors, so it's their cross product; any point on the first line then fixes the plane's position.

   Step-by-Step Solution

   1. Line 1: point $(1,2,1)$, direction $\bar d_1=(1,-1,2)$. Line 2 direction: $\bar d_2=(-1,2,1)$.
   2. Normal $\bar n = \bar d_1\times\bar d_2 = \begin{vmatrix}\bar i&\bar j&\bar k\\1&-1&2\\-1&2&1\end{vmatrix}$.
   3. $\bar i$-component: $(-1)(1)-(2)(2)=-1-4=-5$. $\bar j$-component: $-[(1)(1)-(2)(-1)]=-(1+2)=-3$. $\bar k$-component: $(1)(2)-(-1)(-1)=2-1=1$.
   4. So $\bar n=(-5,-3,1)$.
   5. Plane through $(1,2,1)$: $-5(x-1)-3(y-2)+1(z-1)=0 \Rightarrow -5x-3y+z+10=0 \Rightarrow 5x+3y-z=10$. …

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6. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.$A = (-2, 2, 3), B = (13, -3, 13)$ are two points and P is a variable point such that $PA : PB = 2 : 3$. If P lies on the curve $x^2 + y^2 + z^2 + ux + vy + wz - 247 = 0$, then $u + v + w =$ (A) $24$ (B) $25$ (C) $26$ (D) $27$

   ›Reveal solutionSolution

   The Apollonius locus for $PA:PB=2:3$ is the sphere $9PA^2=4PB^2$; expanding gives

   $x^2+y^2+z^2+28x-12y+10z-247=0$, so $u+v+w=28-12+10=26$.

   Concept and Intuition

   For a fixed ratio $k=PA:PB$ with $k\neq1$, the locus of $P$ is a sphere (the 3D Apollonius

   circle), obtained by squaring the ratio: $PA^2/PB^2=k^2$, i.e. (here) $9PA^2=4PB^2$. Expanding

   this always produces a genuine sphere equation (coefficients of $x^2,y^2,z^2$ equal), which we

   can directly compare to the given form to read off $u,v,w$.

   Step-by-Step Solution

   1. $PA:PB=2:3 \Rightarrow 3PA=2PB \Rightarrow 9PA^2=4PB^2$.
   2. $PA^2=(x+2)^2+(y-2)^2+(z-3)^2$, $PB^2=(x-13)^2+(y+3)^2+(z-13)^2$.
   3. Expand $9PA^2$:

   $9\big[x^2+4x+4+y^2-4y+4+z^2-6z+9\big]=9x^2+9y^2+9z^2+36x-36y-54z+153$

   4. Expand $4PB^2$:

   $4\big[x^2-26x+169+y^2+6y+9+z^2-26z+169\big]=4x^2+4y^2+4z^2-104x+24y-104z+1388$

   5. Set $9PA^2-4PB^2=0$: $5x^2+5y^2+5z^2+140x-60y+50z-1235=0$ …

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7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.An angle between the plane $x + y + z = 5$ and the line $\dfrac{x - 16}{0} = \dfrac{y - 0}{-1} = \dfrac{z + 47}{4}$ is (A) $\mathrm{Sin}^{-1}\left(\dfrac{3}{\sqrt{17}}\right)$ (B) $\mathrm{Sin}^{-1}\left(\sqrt{\dfrac{3}{17}}\right)$ (C) $\mathrm{Cos}^{-1}\left(\sqrt{\dfrac{3}{17}}\right)$ (D) $\mathrm{Sin}^{-1}\left(\dfrac{5}{\sqrt{17}}\right)$

   ›Reveal solutionSolution

   The angle between a line and a plane uses the SINE formula with the line's direction and the

   plane's normal: here $\sin\theta=\sqrt{3/17}$, so $\theta=\sin^{-1}\sqrt{3/17}$.

   Concept and Intuition

   For a line with direction ratios $(l,m,n)$ and a plane with normal $(a,b,c)$, the angle $\theta$

   between the LINE and the PLANE (not the normal) satisfies

   $\sin\theta=\frac{|al+bm+cn|}{\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}$

   — it's a sine (not cosine) because the line-plane angle is complementary to the line-normal angle.

   Step-by-Step Solution

   1. Line: $\dfrac{x-16}{0}=\dfrac{y-0}{-1}=\dfrac{z+47}{4}$, direction ratios $(0,-1,4)$.
   2. Plane: $x+y+z=5$, normal $(1,1,1)$.
   3. Dot product: $(1)(0)+(1)(-1)+(1)(4)=0-1+4=3$.
   4. Magnitudes: $|\vec n|=\sqrt{1+1+1}=\sqrt3$, $|\vec d|=\sqrt{0+1+16}=\sqrt{17}$. …

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8. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If the angle between the planes $\lambda x - 2y + 3z + 1 = 0$ and $2x + 3y - \lambda z + \lambda = 0$ is $\cos^{-1}\left(\dfrac{12}{49}\right)$ and $\lambda \in \mathbb{Z}$ then the sum of the perpendicular distances from the origin to these planes is (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Find the integer $\lambda$ from the given angle between the two planes, then compute and add the two perpendicular distances from the origin.

   Concept and Intuition

   The angle between two planes is the angle between their normal vectors: $\cos\theta=\dfrac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}$. Once the planes are pinned down numerically, the distance of the origin from a plane $ax+by+cz+d=0$ is simply $\dfrac{|d|}{\sqrt{a^2+b^2+c^2}}$.

   Step-by-Step Solution

   1. Normals: $\vec n_1=(\lambda,-2,3)$ and $\vec n_2=(2,3,-\lambda)$.
   2. $\vec n_1\cdot\vec n_2=2\lambda-6-3\lambda=-(\lambda+6)$.
   3. $|\vec n_1|=\sqrt{\lambda^2+4+9}=\sqrt{\lambda^2+13}$ and $|\vec n_2|=\sqrt{4+9+\lambda^2}=\sqrt{\lambda^2+13}$ — equal magnitudes.
   4. So $\cos\theta=\dfrac{|\lambda+6|}{\lambda^2+13}=\dfrac{12}{49}$, i.e. $49|\lambda+6|=12\lambda^2+156$. …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If the four points $(6, 2, 4)$, $(1, 3, 5)$, $(1, -2, 3)$ and $(6, k, 2)$ are coplanar, then $k =$ (A) $-5$ (B) $4$ (C) $-3$ (D) $1$

   ›Reveal solutionSolution

   Four points are coplanar exactly when the scalar triple product of the three edge vectors from a common vertex is zero; solving gives $k=-3$.

   Concept and Intuition

   Four points $P_1,P_2,P_3,P_4$ are coplanar iff the vectors $\vec{P_1P_2}$, $\vec{P_1P_3}$, $\vec{P_1P_4}$ are linearly dependent, i.e. their scalar triple product (the $3\times3$ determinant with these as rows) is zero — geometrically, the tetrahedron they'd otherwise form has zero volume.

   Step-by-Step Solution

   1. Let $P_1=(6,2,4)$, $P_2=(1,3,5)$, $P_3=(1,-2,3)$, $P_4=(6,k,2)$.
   2. $\vec v_1=P_2-P_1=(-5,1,1)$; $\vec v_2=P_3-P_1=(-5,-4,-1)$; $\vec v_3=P_4-P_1=(0,k-2,-2)$.
   3. Set up the determinant:

   $\begin{vmatrix}-5&1&1\\-5&-4&-1\\0&k-2&-2\end{vmatrix}=0$

   4. Expand along row 1: $-5[(-4)(-2)-(-1)(k-2)] - 1[(-5)(-2)-(-1)(0)] + 1[(-5)(k-2)-(-4)(0)]$.
   5. First bracket: $8-(-(k-2))=8+k-2=k+6$, times $-5$: $-5k-30$.
   6. Second term: $-1[10-0]=-10$.
   7. Third term: $1[-5(k-2)-0]=-5k+10$. …

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10. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.A plane $\pi$ given by $ax+by+11z+d=0$ is perpendicular to the planes $2x-3y+z=4$, $3x+y-z=5$ and the perpendicular distance from the origin to the plane $\pi$ is $\sqrt6$ units. If all the intercepts made by the plane $\pi$ on the coordinate axes are positive, then $d=$ (A) $ab$ (B) $-2ab$ (C) $4ab$ (D) $-3ab$

    ›Reveal solutionSolution

    Two perpendicularity conditions fix $a,b$; the distance condition fixes $|d|$; the positive-intercepts condition fixes the sign of $d$. The result matches $-3ab$.

    Concept and Intuition

    Two planes are perpendicular exactly when their normal vectors are perpendicular (dot product zero). This gives two linear equations in $a,b$ (with the $z$-coefficient 11 already fixed). The distance-from-origin condition then pins down $|d|$ using the standard point-to-plane distance formula. Finally, intercepts on the axes are found by setting two of the three coordinates to zero at a time; requiring all three intercepts positive fixes the sign of $d$ relative to the signs of $a,b,11$.

    Step-by-Step Solution

    1. Plane $\pi: ax+by+11z+d=0$, normal $(a,b,11)$.
    2. Perpendicular to $2x-3y+z=4$ (normal $(2,-3,1)$): $2a-3b+11=0$.
    3. Perpendicular to $3x+y-z=5$ (normal $(3,1,-1)$): $3a+b-11=0 \Rightarrow b=11-3a$.
    4. Substitute into step 2: $2a-3(11-3a)+11=0 \Rightarrow 2a-33+9a+11=0 \Rightarrow 11a-22=0\Rightarrow a=2$.
    5. $b=11-3(2)=5$. So $\pi: 2x+5y+11z+d=0$.
    6. Distance from origin: $\dfrac{|0+0+0+d|}{\sqrt{2^2+5^2+11^2}}=\dfrac{|d|}{\sqrt{4+25+121}}=\dfrac{|d|}{\sqrt{150}}=\sqrt6$. $|d|=\sqrt6\cdot\sqrt{150}=\sqrt{900}=30$. …

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11. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.If the image of the point A(1, 1, 1) with respect to the plane $4x + 2y + 4z + 1 = 0$ is $B(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma =$ (A) $-2$ (B) $-\dfrac{28}{9}$ (C) $\dfrac{55}{36}$ (D) $\dfrac{35}{16}$

    ›Reveal solutionSolution

    Standard reflection-in-a-plane formula. Answer: $\alpha+\beta+\gamma=-\dfrac{28}{9}$.

    Concept and Intuition

    The image (reflection) of a point $(x_1,y_1,z_1)$ in the plane $ax+by+cz+d=0$ lies along the normal direction $(a,b,c)$ through the point, at twice the signed distance to the plane on the opposite side:

    $\frac{x'-x_1}{a}=\frac{y'-y_1}{b}=\frac{z'-z_1}{c}=-\frac{2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}.$

    Step-by-Step Solution

    1. Here $a=4,\ b=2,\ c=4,\ d=1$ and $(x_1,y_1,z_1)=(1,1,1)$.
    2. $ax_1+by_1+cz_1+d = 4+2+4+1=11$.
    3. $a^2+b^2+c^2=16+4+16=36$.
    4. Factor $t=-\dfrac{2(11)}{36}=-\dfrac{11}{18}$.
    5. $\alpha = x_1+at = 1+4\left(-\dfrac{11}{18}\right)=1-\dfrac{44}{18}=1-\dfrac{22}{9}=-\dfrac{13}{9}$.
    6. $\beta = y_1+bt = 1+2\left(-\dfrac{11}{18}\right)=1-\dfrac{22}{18}=1-\dfrac{11}{9}=-\dfrac29$. …

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12. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.The points $A(-1, 2, 3)$, $B(2, -3, 1)$, $C(3, 1, -2)$ (A) are collinear (B) form an isosceles triangle (C) form a right angled triangle (D) form a scalene triangle

    ›Reveal solutionSolution

    Compute all three squared side lengths of triangle $ABC$ and test them for equality (isosceles), a Pythagorean relation (right angle), and proportionality of two sides (collinearity) — none hold, so it's scalene.

    Concept and Intuition

    Given three points in space, the quickest classification route is via squared distances (avoiding square roots): compare all three pairwise squared distances for equality (isosceles/equilateral) and for a Pythagorean sum (right angle), and check if any two connecting vectors are parallel (collinearity).

    Step-by-Step Solution

    1. $A(-1,2,3)$, $B(2,-3,1)$, $C(3,1,-2)$.
    2. $\vec{AB}=(2-(-1),\,-3-2,\,1-3)=(3,-5,-2)$, so $AB^2=9+25+4=38$.
    3. $\vec{BC}=(3-2,\,1-(-3),\,-2-1)=(1,4,-3)$, so $BC^2=1+16+9=26$.
    4. $\vec{CA}=(-1-3,\,2-1,\,3-(-2))=(-4,1,5)$, so $CA^2=16+1+25=42$.
    5. Isosceles check: $38,26,42$ are all distinct, so no two sides are equal.
    6. Right-angle check: $38+26=64\ne42$; $26+42=68\ne38$; $38+42=80\ne26$ — no pair of squared sides sums to the third, so there is no right angle. …

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