Q.The number of vectors of unit length perpendicular to the vectors $\vec{a}=2\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$ is
(A) one
(B) two
(C) three
(D) infinite

Showing the 12 most recent of 25 on this concept.

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

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A unit vector that is perpendicular to the vector $2\bar{i} - \bar{j} + 2\bar{k}$ and coplanar with the vectors $\bar{i} + \bar{j} - \bar{k}$ and $2\bar{i} + 2\bar{j} - \bar{k}$ is (A) $\dfrac{\bar{i} + 2\bar{j} + \bar{k}}{\sqrt{6}}$ (B) $\dfrac{3\bar{i} + 2\bar{j} - 2\bar{k}}{\sqrt{17}}$ (C) $\dfrac{2\bar{i} + 2\bar{j} - \bar{k}}{3}$ (D) $\dfrac{3\bar{i} + 2\bar{j} + 2\bar{k}}{\sqrt{17}}$

   ›Reveal solutionSolution

   Write the general coplanar combination of the two given vectors as $a\,p+b\,q$, impose perpendicularity to the third vector to pin down $a=0$, then normalize the resulting direction.

   Concept and Intuition

   "Coplanar with $p$ and $q$" means the target vector is some linear combination $a\,p+b\,q$ (this spans exactly the plane through the origin containing both). Imposing perpendicularity to a third given vector is then just one linear equation in $a,b$ — it typically forces a ratio (or here, forces one coefficient to vanish entirely), collapsing the family to a single direction, which we then normalize to a unit vector.

   Step-by-Step Solution

   1. General coplanar vector: $v=a(\bar i+\bar j-\bar k)+b(2\bar i+2\bar j-\bar k)=(a+2b)\bar i+(a+2b)\bar j+(-a-b)\bar k$.
   2. Require $v\perp(2\bar i-\bar j+2\bar k)$: $2(a+2b)-1(a+2b)+2(-a-b)=0$.
   3. Simplify: $(a+2b)(2-1)+2(-a-b)=(a+2b)-2a-2b=-a$.
   4. So the condition reduces to $-a=0\Rightarrow a=0$.
   5. With $a=0$: $v=b(2\bar i+2\bar j-\bar k)$, i.e. $v$ is parallel to $2\bar i+2\bar j-\bar k$.
   6. Magnitude of $2\bar i+2\bar j-\bar k$ is $\sqrt{4+4+1}=3$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The vector of magnitude 2 lying in the plane of $\bar{a} = 2\bar{i} - \bar{j} + \bar{k}$ and $\bar{b} = \bar{i} + 3\bar{j} - 5\bar{k}$ and perpendicular to the vector $\bar{c} = \bar{i} + \bar{j} + \bar{k}$ is (A) $\dfrac{\sqrt{2}}{\sqrt{61}}(4\bar{i}+5\bar{j}-9\bar{k})$ (B) $\dfrac{\sqrt{2}}{\sqrt{9}}(2\bar{i}+3\bar{j}-5\bar{k})$ (C) $\dfrac{\sqrt{2}}{\sqrt{31}}(\bar{i}+5\bar{j}-6\bar{k})$ (D) $\dfrac{\sqrt{2}}{\sqrt{13}}(-\bar{i}-3\bar{j}+4\bar{k})$

   ›Reveal solutionSolution

   This tests writing a vector "in the plane of $\bar a,\bar b$" as a linear combination $\alpha\bar a+\beta\bar b$, using perpendicularity to $\bar c$ to pin the ratio $\alpha:\beta$, and finally scaling the resulting direction to the required magnitude.

   Concept and Intuition

   Every vector lying in the plane spanned by $\bar a$ and $\bar b$ is some linear combination $\alpha\bar a+\beta\bar b$ — that's what "lying in the plane" means. The extra condition (perpendicular to $\bar c$) gives one linear equation in $\alpha,\beta$, which fixes their ratio (the direction is determined up to an overall scale). The magnitude condition then fixes that scale.

   Step-by-Step Solution

   1. Let $\bar d=\alpha\bar a+\beta\bar b$ for some scalars $\alpha,\beta$ (this covers every vector in the plane of $\bar a,\bar b$).
   2. Require $\bar d\cdot\bar c=0$: $\alpha(\bar a\cdot\bar c)+\beta(\bar b\cdot\bar c)=0$.
   3. $\bar a\cdot\bar c=(2)(1)+(-1)(1)+(1)(1)=2-1+1=2$. $\bar b\cdot\bar c=(1)(1)+(3)(1)+(-5)(1)=1+3-5=-1$.
   4. So $2\alpha-\beta=0\Rightarrow\beta=2\alpha$. Taking $\alpha=1,\beta=2$: direction $=\bar a+2\bar b=(2+2,\,-1+6,\,1-10)=(4,5,-9)$.
   5. Magnitude of this direction: $\sqrt{4^2+5^2+(-9)^2}=\sqrt{16+25+81}=\sqrt{122}$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.$\bar{a} = \bar{i} - \bar{j} + \bar{k}$, $\bar{b} = 2\bar{i} + \bar{j} + \bar{k}$ are two vectors and $\bar{c}$ is a unit vector lying in the plane of $\bar{a}$ and $\bar{b}$. If $\bar{c}$ is perpendicular to $\bar{b}$ then $\bar{c}.(\bar{i}+\bar{j}+2\bar{k}) =$ (A) $0$ (B) $5$ (C) $\dfrac{1}{\sqrt{21}}$ (D) $\dfrac{2}{\sqrt{21}}$

   ›Reveal solutionSolution

   This tests finding a unit vector coplanar with two given vectors and perpendicular to one of them; the required dot product works out to $\dfrac{1}{\sqrt{21}}$.

   Concept and Intuition

   Any vector in the plane spanned by $\bar a$ and $\bar b$ can be written as a linear combination $m\bar a+n\bar b$. Imposing perpendicularity to $\bar b$ gives one linear equation in $m,n$, pinning down the direction of $\bar c$ up to a scalar (which is then fixed by the unit-length condition).

   Step-by-Step Solution

   1. Let $\bar c=m\bar a+n\bar b$ where $\bar a=(1,-1,1),\ \bar b=(2,1,1)$.
   2. $\bar c\cdot\bar b=0\Rightarrow m(\bar a\cdot\bar b)+n(\bar b\cdot\bar b)=0$.
   3. $\bar a\cdot\bar b=1(2)+(-1)(1)+1(1)=2-1+1=2$. $\bar b\cdot\bar b=4+1+1=6$.
   4. So $2m+6n=0\Rightarrow m=-3n$.
   5. $\bar c\parallel -3n\bar a+n\bar b=n(-3\bar a+\bar b)=n\big((-3,3,-3)+(2,1,1)\big)=n(-1,4,-2)$.
   6. Direction vector $(-1,4,-2)$ has magnitude $\sqrt{1+16+4}=\sqrt{21}$, so the unit vector is $\pm\dfrac{(-1,4,-2)}{\sqrt{21}}$. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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: $\bar d = (3t,-5t,4t) = t(3,-5,4)$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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: $l-m=1$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $\bar a\times(\bar b\times\bar c)=\begin{vmatrix}\bar i&\bar j&\bar k\\2&3&0\\12&20&-15\end{vmatrix}$ …

   🔒 Start your 14-day free trial to see the full solution →
8. 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: $(1)(-1)+(1)(-1)+(1)(2) = -1-1+2 = 0$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $=3^2-4(2)(5)=9-40=-31<0$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $|\vec a|=2$, $|\vec b|=3$: $4 = 9t^2 \Rightarrow t^2=\dfrac49$. …

    🔒 Start your 14-day free trial to see the full solution →
11. 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 $w$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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. $|\vec r|^2=\lambda^2(846)=94\Rightarrow\lambda^2=\dfrac{94}{846}=\dfrac19\Rightarrow\lambda=\pm\dfrac13$. …

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