Q.Prove that $(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2$, if and only if $\vec{a},\vec{b}$ are perpendicular, given $\vec{a}\neq\vec{0},\vec{b}\neq\vec{0}$.

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $\vec a$ and $\vec b$ are two vectors such that $|\vec a| = 2$, $|\vec b| = 3$ and $\vec a + t\vec b$ and $\vec a - t\vec b$ are perpendicular, where 't' is a positive scalar, then (A) $t = \pm\dfrac{2}{3}$ (B) $t = \dfrac{4}{9}$ (C) $t = \dfrac{2}{3}$ (D) $t = \dfrac{2}{9}$

   ›Reveal solutionSolution

   Perpendicularity of $\vec a+t\vec b$ and $\vec a-t\vec b$ forces $|\vec a|^2=t^2|\vec b|^2$, giving the positive value $t=2/3$.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product is zero. Expanding $(\vec a+t\vec b)\cdot(\vec a-t\vec b)$ using the distributive property of the dot product collapses to a simple difference of squared magnitudes, since $\vec a\cdot\vec b$ cancels.

   Step-by-Step Solution

   1. $(\vec a+t\vec b)\cdot(\vec a-t\vec b) = \vec a\cdot\vec a - t\,\vec a\cdot\vec b + t\,\vec b\cdot\vec a - t^2\,\vec b\cdot\vec b = |\vec a|^2 - t^2|\vec b|^2$.
   2. Setting this to zero (perpendicularity): $|\vec a|^2 = t^2|\vec b|^2$.
   3. Substitute $|\vec a|=2$, $|\vec b|=3$: $4 = 9t^2 \Rightarrow t^2=\dfrac49$. …

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2. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.$\bar{a},\bar{b},\bar{c}$ are unit vectors. If $\bar{a},\bar{b}$ are perpendicular vectors, $(\bar{a}-\bar{c}).(\bar{b}+\bar{c})=0$ and $\bar{c}=l\bar{a}+m\bar{b}+n(\bar{a}\times\bar{b})$; (l, m, n are scalars), then $n^2=$ (A) $l^2+m^2$ (B) $-2lm$ (C) $2l-2m$ (D) $lm+l+m$

   ›Reveal solutionSolution

   Because $\bar a,\bar b,\bar a\times\bar b$ form an orthonormal triad, decomposing $\bar c$ in this basis and using the given perpendicularity condition shows $n^2=-2lm$.

   Concept and Intuition

   When $\bar a$ and $\bar b$ are perpendicular unit vectors, $\bar a\times\bar b$ is automatically a unit vector too (since $|\bar a\times\bar b|=|\bar a||\bar b|\sin 90°=1$) and is perpendicular to both $\bar a$ and $\bar b$. So $\{\bar a,\bar b,\bar a\times\bar b\}$ is an orthonormal basis — any vector's components along these three directions are just its dot products with each, and its squared magnitude is simply the sum of squared components (Pythagoras in 3D).

   Step-by-Step Solution

   1. Since $\bar a\perp\bar b$ and both are unit vectors, $\bar a.\bar b=0$ and $\{\bar a,\bar b,\bar a\times\bar b\}$ is orthonormal.
   2. Expand $(\bar a-\bar c).(\bar b+\bar c)=0$: $\bar a.\bar b+\bar a.\bar c-\bar c.\bar b-\bar c.\bar c=0$.
   3. Since $\bar a.\bar b=0$ and $\bar c.\bar c=|\bar c|^2=1$ (unit vector): $\bar a.\bar c-\bar b.\bar c-1=0 \Rightarrow \bar a.\bar c-\bar b.\bar c=1$.
   4. Given $\bar c=l\bar a+m\bar b+n(\bar a\times\bar b)$, dot with $\bar a$: $\bar a.\bar c=l(\bar a.\bar a)+m(\bar a.\bar b)+n\cdot\bar a.(\bar a\times\bar b)=l(1)+m(0)+n(0)=l$ (since $\bar a.(\bar a\times\bar b)=0$, a vector is always perpendicular to a cross product it's part of).
   5. Similarly, dot with $\bar b$: $\bar b.\bar c=l(\bar b.\bar a)+m(\bar b.\bar b)+n\cdot\bar b.(\bar a\times\bar b)=0+m(1)+0=m$.
   6. From step 3: $l-m=1$. …

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3. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If ABCD is a cyclic quadrilateral with R as the radius of the circumcircle and $(AB)^2 + (CD)^2 = 4R^2$ then (A) $\bar{b}.\bar{c} - \bar{a}.\bar{d} = 0$ (B) $\bar{a}.\bar{c} - \bar{b}.\bar{d} = 0$ (C) $\bar{a}.\bar{b} + \bar{c}.\bar{d} = 0$ (D) $\bar{a}.\bar{c} + \bar{b}.\bar{d} = 0$

   ›Reveal solutionSolution

   Expressing each chord-length in terms of position vectors from the circumcenter turns the given length condition directly into a dot-product identity — the answer is (C).

   Concept and Intuition

   For points on a circle of radius $R$ centered at the origin, the squared distance between two points $\bar p,\bar q$ on the circle is $|\bar q-\bar p|^2 = |\bar q|^2+|\bar p|^2-2\bar p\cdot\bar q = 2R^2-2\bar p\cdot\bar q$, since $|\bar p|=|\bar q|=R$.

   Step-by-Step Solution

   1. Let $\bar a,\bar b,\bar c,\bar d$ be position vectors of $A,B,C,D$ from the circumcenter, so $|\bar a|=|\bar b|=|\bar c|=|\bar d|=R$.
   2. $AB^2 = |\bar b-\bar a|^2 = |\bar a|^2+|\bar b|^2-2\bar a\cdot\bar b = 2R^2-2\bar a\cdot\bar b$.
   3. Similarly, $CD^2 = 2R^2-2\bar c\cdot\bar d$.
   4. Given $AB^2+CD^2=4R^2$: $(2R^2-2\bar a\cdot\bar b)+(2R^2-2\bar c\cdot\bar d) = 4R^2$. …

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4. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $\vec a = \dfrac{3}{2}\hat k$, $\vec b = \dfrac{2\hat i + 2\hat j - \hat k}{2}$, then angle between $\vec a + \vec b$ and $\vec a - \vec b$ is (A) $45^\circ$ (B) $90^\circ$ (C) $30^\circ$ (D) $60^\circ$

   ›Reveal solutionSolution

   Computing $(\vec a+\vec b)\cdot(\vec a-\vec b)$ gives zero, so the two vectors are perpendicular.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product vanishes. Rather than compute the angle via magnitudes and cosine, it's fastest to just test $(\vec a+\vec b)\cdot(\vec a-\vec b) = |\vec a|^2-|\vec b|^2$ or, more generally here, expand directly since $\vec a,\vec b$ aren't simply given by magnitude alone (they have specific components).

   Step-by-Step Solution

   1. $\vec a = \dfrac32\hat k = (0,0,\tfrac32)$.
   2. $\vec b = \dfrac{2\hat i+2\hat j-\hat k}{2} = \hat i+\hat j-\dfrac12\hat k = (1,1,-\tfrac12)$.
   3. $\vec a+\vec b = (1,1,\tfrac32-\tfrac12) = (1,1,1)$.
   4. $\vec a-\vec b = (-1,-1,\tfrac32+\tfrac12) = (-1,-1,2)$.
   5. Dot product: $(1)(-1)+(1)(-1)+(1)(2) = -1-1+2 = 0$. …

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5. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.Let $u = 2i + 3j + k$, $v = -3i + 2j$ and $w = i - j + 4k$. Then which of the following statement is true? (A) $u$ is perpendicular to $v$ but not $w$ (B) $v$ is perpendicular to $w$ but not $u$ (C) $w$ is perpendicular to $u$ but not $v$ (D) $u$ is perpendicular to both $v$ and $w$

   ›Reveal solutionSolution

   Direct dot products show $u\cdot v=0$ (perpendicular) and $u\cdot w=3\ne0$ (not perpendicular), matching option (A).

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product is zero. Checking each pair's dot product directly settles every option — no need for angle or cross-product computation.

   Step-by-Step Solution

   1. $u\cdot v = (2)(-3)+(3)(2)+(1)(0) = -6+6+0 = 0$ — so $u\perp v$.
   2. $u\cdot w = (2)(1)+(3)(-1)+(1)(4) = 2-3+4 = 3 \ne 0$ — so $u$ is not perpendicular to $w$.
   3. For completeness, $v\cdot w=(-3)(1)+(2)(-1)+(0)(4)=-3-2+0=-5\ne0$ — $v$ is also not perpendicular to $w$. …

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6. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.The number of vectors of unit length perpendicular to the two vectors $\vec{a} = (1,1,0)$ and $\vec{b} = (0,1,1)$ is (A) $1$ (B) $2$ (C) $3$ (D) Infinite

   ›Reveal solutionSolution

   The cross product gives one direction perpendicular to both vectors, and its two unit multiples ($+$ and $-$) are the only unit vectors satisfying the condition.

   Concept and Intuition

   Any vector perpendicular to two given non-parallel vectors must be a scalar multiple of their cross product; normalizing gives exactly two opposite unit vectors.

   Step-by-Step Solution

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

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7. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.The set of real values of λ for which the vectors $\lambda\vec{i}-3\vec{j}+5\vec{k}$ and $2\lambda\vec{i}-\lambda\vec{j}+\vec{k}$ are perpendicular to each other is (A) $\{0, 1\}$ (B) $\{-2\}$ (C) $\{2, -1\}$ (D) $\varphi$

   ›Reveal solutionSolution

   Perpendicular vectors have zero dot product; the resulting quadratic in $\lambda$ has no real roots, so the answer set is empty.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product vanishes. Setting up that equation converts a geometry condition into an algebraic one in $\lambda$.

   Step-by-Step Solution

   1. The vectors are $\vec u=(\lambda,-3,5)$ and $\vec v=(2\lambda,-\lambda,1)$.
   2. Perpendicularity: $\vec u\cdot\vec v=0$: $\ \lambda(2\lambda)+(-3)(-\lambda)+5(1)=0$.
   3. Simplify: $2\lambda^2+3\lambda+5=0$.
   4. Discriminant $=3^2-4(2)(5)=9-40=-31<0$. …

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8. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $\bar{a}=2\bar{i}+3\bar{j}, \bar{b}=3\bar{j}+4\bar{k}$ and $\bar{c}=5\bar{i}+4\bar{k}$ are three vectors, then a vector which is perpendicular to $\bar{a}$ and $\bar{b}\times\bar{c}$ is (A) $45\bar{i}-30\bar{j}+15\bar{k}$ (B) $3\bar{i}-2\bar{j}+\bar{k}$ (C) $-30\bar{i}+20\bar{j}+4\bar{k}$ (D) $-45\bar{i}+30\bar{j}+4\bar{k}$

   ›Reveal solutionSolution

   This tests the vector-triple-product idea: a vector perpendicular to both $\bar a$ and $\bar b\times\bar c$ is simply $\bar a\times(\bar b\times\bar c)$.

   Concept and Intuition

   The cross product of any two vectors is perpendicular to both of them. So if we want a single vector perpendicular to $\bar a$ AND to $\bar b\times\bar c$, the natural candidate is $\bar a\times(\bar b\times\bar c)$ — it is perpendicular to $\bar a$ by definition of cross product, and perpendicular to $\bar b\times\bar c$ for the same reason. No need to invoke the full triple-product expansion formula; we just compute it directly.

   Step-by-Step Solution

   1. Given $\bar a=2\bar i+3\bar j+0\bar k$, $\bar b=0\bar i+3\bar j+4\bar k$, $\bar c=5\bar i+0\bar j+4\bar k$.
   2. Compute $\bar b\times\bar c=\begin{vmatrix}\bar i&\bar j&\bar k\\0&3&4\\5&0&4\end{vmatrix}$ $=\bar i(3\cdot4-4\cdot0)-\bar j(0\cdot4-4\cdot5)+\bar k(0\cdot0-3\cdot5)=12\bar i+20\bar j-15\bar k$.
   3. Compute $\bar a\times(\bar b\times\bar c)=\begin{vmatrix}\bar i&\bar j&\bar k\\2&3&0\\12&20&-15\end{vmatrix}$ …

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9. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.Let $\vec a=2\vec i-3\vec j-5\vec k$ and $\vec b=3\vec i+2\vec j-5\vec k$ be two vectors and $\vec r$ be a vector in the plane of $\vec a$ and $\vec b$. If $\vec r$ is orthogonal to the vector $5\vec i-2\vec j+3\vec k$ and the magnitude of $\vec r$ is $\sqrt{94}$, then $|\vec r \cdot \vec b| =$ (A) $36$ (B) $38$ (C) $42$ (D) $46$

   ›Reveal solutionSolution

   Since $\vec r$ is in the plane of $\vec a,\vec b$ and perpendicular to $\vec n$, it must be parallel to $(\vec a\times\vec b)\times\vec n$; scaling this to the given magnitude $\sqrt{94}$ and dotting with $\vec b$ gives $|\vec r\cdot\vec b|=46$.

   Concept and Intuition

   Two conditions pin down $\vec r$'s direction uniquely (up to sign and scale): (1) $\vec r$ lies in the plane of $\vec a,\vec b$, meaning $\vec r\perp N$ where $N=\vec a\times\vec b$ is the plane's normal; (2) $\vec r\perp\vec n$ (given). A vector perpendicular to both $N$ and $\vec n$ must be parallel to $N\times\vec n$.

   Step-by-Step Solution

   1. $\vec a=(2,-3,-5)$, $\vec b=(3,2,-5)$. Compute $N=\vec a\times\vec b$: $N_i=(-3)(-5)-(-5)(2)=15+10=25$ $N_j=-[(2)(-5)-(-5)(3)]=-[-10+15]=-5$ $N_k=(2)(2)-(-3)(3)=4+9=13$ So $N=(25,-5,13)$.
   2. $\vec n=(5,-2,3)$. Compute $N\times\vec n$: $i$: $(-5)(3)-(13)(-2)=-15+26=11$ $j$: $-[(25)(3)-(13)(5)]=-[75-65]=-10$ $k$: $(25)(-2)-(-5)(5)=-50+25=-25$ So $N\times\vec n=(11,-10,-25)$.
   3. $\vec r=\lambda(11,-10,-25)$ for some scalar $\lambda$. $|N\times\vec n|^2=121+100+625=846$.
   4. $|\vec r|^2=\lambda^2(846)=94\Rightarrow\lambda^2=\dfrac{94}{846}=\dfrac19\Rightarrow\lambda=\pm\dfrac13$. …

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10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.Let $\bar{a} = 3\bar{i} - \bar{j} - \bar{k}$, $\bar{b} = \bar{i} + \bar{j} - 2\bar{k}$ and $\bar{c} = 2\bar{i} + 2\bar{j} + \bar{k}$. Let $\bar{d}$ be a vector such that $|\bar{d}| = \sqrt{2}$ units. If the vector $\bar{d}$ is coplanar with $\bar{a}, \bar{b}$ and perpendicular to $\bar{c}$, then $\bar{d} =$ (A) $\pm\dfrac{1}{5}(3\bar{i} - 5\bar{j} + 4\bar{k})$ (B) $\pm\dfrac{1}{5}(-4\bar{i} + 5\bar{j} - 3\bar{k})$ (C) $\pm\dfrac{1}{5}(3\bar{i} + 5\bar{j} - 4\bar{k})$ (D) $\pm\dfrac{1}{5}(-3\bar{i} + 5\bar{j} + 4\bar{k})$

    ›Reveal solutionSolution

    $\bar d$ coplanar with $\bar a,\bar b$ means $\bar d=x\bar a+y\bar b$; perpendicularity to $\bar c$ fixes the ratio $x:y$; the given magnitude fixes the scale. The answer is (A).

    Concept and Intuition

    "Coplanar with $\bar a,\bar b$" means $\bar d$ lies in the plane spanned by $\bar a$ and $\bar b$, so it can be written as a linear combination $\bar d=x\bar a+y\bar b$ for some scalars $x,y$ (this is exactly what "coplanar with two given vectors, through the origin" means). The perpendicularity condition $\bar d\cdot\bar c=0$ then gives one constraint relating $x$ and $y$, so $\bar d$ is pinned down up to a single scalar multiple — which the given magnitude $|\bar d|=\sqrt2$ finally fixes (up to sign, since both directions along that line satisfy all the stated conditions).

    Step-by-Step Solution

    1. Given $\bar a=(3,-1,-1)$, $\bar b=(1,1,-2)$, $\bar c=(2,2,1)$.
    2. Since $\bar d$ is coplanar with $\bar a,\bar b$, write $\bar d = x\bar a + y\bar b = (3x+y,\,-x+y,\,-x-2y)$.
    3. Perpendicularity to $\bar c$: $\bar d\cdot\bar c=0$:

    $2(3x+y)+2(-x+y)+1(-x-2y)=0$

    $6x+2y-2x+2y-x-2y=0 \implies 3x+2y=0 \implies y=-\tfrac{3x}{2}.$

    4. Substitute back:

    $\bar d = \left(3x-\tfrac{3x}{2},\ -x-\tfrac{3x}{2},\ -x+3x\right) = \left(\tfrac{3x}{2},\,-\tfrac{5x}{2},\,2x\right).$

    Let $x=2t$ to clear fractions: $\bar d = (3t,-5t,4t) = t(3,-5,4)$. …

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11. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.In $\triangle ABC$, if $\overline{AB} = 2\bar{i} - \bar{j} + 2\bar{k}$ and $\overline{AC} = 3\bar{i} - 3\bar{j} + 4\bar{k}$, then that triangle $ABC$ is (A) an equilateral triangle (B) a right angled triangle (C) an isosceles triangle (D) a scalene triangle

    ›Reveal solutionSolution

    Compute all three side lengths of $\triangle ABC$ from $\overline{AB}$ and $\overline{AC}$ (with $\overline{BC}=\overline{AC}-\overline{AB}$) and classify by comparing them. Answer: isosceles triangle.

    Concept and Intuition

    Given two sides of a triangle as vectors from a common vertex, the third side is simply their difference (vector subtraction along the triangle). Once all three side lengths are known, classifying the triangle (scalene / isosceles / equilateral / right-angled) is a direct numeric comparison — no need for angles unless a right angle is suspected, in which case the converse of the Pythagorean theorem settles it.

    Step-by-Step Solution

    1. $\overline{AB}=(2,-1,2)$, so $|AB|=\sqrt{2^2+(-1)^2+2^2}=\sqrt{4+1+4}=\sqrt9=3$.
    2. $\overline{AC}=(3,-3,4)$, so $|AC|=\sqrt{9+9+16}=\sqrt{34}$.
    3. $\overline{BC}=\overline{AC}-\overline{AB}=(3-2,-3-(-1),4-2)=(1,-2,2)$, so $|BC|=\sqrt{1+4+4}=\sqrt9=3$.
    4. The three sides are $3, 3, \sqrt{34}$ — two equal sides, so the triangle is isosceles, not equilateral (all three would need to be equal) and not scalene (that needs all different). …

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12. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.Let ABC be an equilateral triangle of side $a$. M and N are two points on the sides AB and AC respectively such that $\overline{AN} = K\overline{AC}$ and $\overline{AB} = 3\overline{AM}$. If the vectors $\overline{BN}$ and $\overline{CM}$ are perpendicular, then $K =$ (A) $\dfrac{1}{5}$ (B) $\dfrac{2}{5}$ (C) $-\dfrac{1}{5}$ (D) $-\dfrac{2}{5}$

    ›Reveal solutionSolution

    Express $\overline{BN}$ and $\overline{CM}$ in terms of the two sides from $A$, use the $60^\circ$ dot product of an equilateral triangle, and set the perpendicularity condition to zero to solve for $K=\dfrac15$.

    Concept and Intuition

    Placing the vertex $A$ at the origin turns every other point into a simple scalar multiple of the two side vectors $\overline{AB}$ and $\overline{AC}$. Perpendicularity of two vectors becomes an algebraic condition: their dot product is zero. For an equilateral triangle, $\overline{AB}.\overline{AC}=a^2\cos 60^\circ=\dfrac{a^2}{2}$.

    Step-by-Step Solution

    1. Let $A$ be the origin, $\bar c=\overline{AB}$, $\bar b=\overline{AC}$, with $|\bar b|=|\bar c|=a$ and $\bar b.\bar c=\dfrac{a^2}{2}$.
    2. Since $\overline{AB}=3\overline{AM}$, $M=\dfrac{\bar c}{3}$. Since $\overline{AN}=K\overline{AC}$, $N=K\bar b$.
    3. $\overline{BN}=N-B=K\bar b-\bar c$, and $\overline{CM}=M-C=\dfrac{\bar c}{3}-\bar b$.
    4. Perpendicularity: $\overline{BN}.\overline{CM}=0$: $(K\bar b-\bar c).\left(\dfrac{\bar c}{3}-\bar b\right)= \dfrac{K}{3}(\bar b.\bar c) - K|\bar b|^2 - \dfrac13|\bar c|^2 + \bar b.\bar c = 0$.
    5. Substitute $|\bar b|^2=|\bar c|^2=a^2$, $\bar b.\bar c=a^2/2$: …

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