Q.If $\vec{a}=5\hat{i}-\hat{j}-3\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}-5\hat{k}$, then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

Showing the 12 most recent of 25 on this concept.

1. 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: $(1)(-1)+(1)(-1)+(1)(2) = -1-1+2 = 0$.
   6. Zero dot product means the vectors are perpendicular, so the angle between them is $90^\circ$.

   Common Mistakes

   • Arithmetic slip when combining the $\hat k$-components of $\vec a$ and $\vec b$.
   • Assuming the identity $(\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2$ applies without checking — it does here since cross terms cancel identically regardless, but direct computation is safest.

   ✓Final answer

   The correct option is (B) — $90^\circ$.

   ANSWER: B

2. 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 $\bar a\times(\bar b\times\bar c)=\begin{vmatrix}\bar i&\bar j&\bar k\\2&3&0\\12&20&-15\end{vmatrix}$ $=\bar i(3\cdot(-15)-0\cdot20)-\bar j(2\cdot(-15)-0\cdot12)+\bar k(2\cdot20-3\cdot12)$ $=\bar i(-45)-\bar j(-30)+\bar k(4)=-45\bar i+30\bar j+4\bar k$.
   4. This matches option (D).

   Common Mistakes

   • Computing $\bar c\times\bar b$ or $(\bar b\times\bar c)\times\bar a$ instead — cross product is anti-commutative, so sign/order matters.
   • Arithmetic slips in the $3\times3$ determinant expansion (missing the $-$ sign on the $\bar j$ cofactor).

   ✓Final answer

   The correct option is (D) — $-45\bar{i}+30\bar{j}+4\bar{k}$.

   ANSWER: D

3. 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. $|\vec r|^2=\lambda^2(846)=94\Rightarrow\lambda^2=\dfrac{94}{846}=\dfrac19\Rightarrow\lambda=\pm\dfrac13$.
   5. $\vec r\cdot\vec b=\lambda\big[11(3)+(-10)(2)+(-25)(-5)\big]=\lambda[33-20+125]=\lambda(138)$.
   6. With $\lambda=\pm\dfrac13$: $\vec r\cdot\vec b=\pm46$. So $|\vec r\cdot\vec b|=46$.

   Common Mistakes

   • Forgetting that "in the plane of $\vec a,\vec b$ AND perpendicular to $\vec n$" forces $\vec r$ along a specific cross-product direction, and instead trying to write $\vec r=x\vec a+y\vec b$ with two unknowns (a valid but far more laborious route).
   • Sign errors in the two successive cross-product computations.

   ✓Final answer

   The correct option is (D) — $46$.

   ANSWER: D

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 $\pm\dfrac{(-1,4,-2)}{\sqrt{21}}$.
   7. Dot with $(1,1,2)$: $(-1)(1)+4(1)+(-2)(2)=-1+4-4=-1$. So $\bar c\cdot(\bar i+\bar j+2\bar k)=\mp\dfrac{1}{\sqrt{21}}$, i.e. magnitude $\dfrac{1}{\sqrt{21}}$.

   Common Mistakes

   • Forgetting that "unit vector in the plane" still leaves a sign ambiguity — the magnitude of the answer is what's pinned down, matching option (C)'s value $\frac{1}{\sqrt{21}}$ rather than the larger $\frac{2}{\sqrt{21}}$ in (D).
   • Computing $\bar a\cdot\bar b$ or $\bar b\cdot\bar b$ incorrectly and getting the wrong ratio $m:n$.

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{\sqrt{21}}$.

   ANSWER: C

5. 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 $|\bar b|^2=|\bar c|^2=a^2$, $\bar b.\bar c=a^2/2$: $\dfrac{K}{3}\cdot\dfrac{a^2}{2} - Ka^2 - \dfrac{a^2}{3} + \dfrac{a^2}{2}=0$.
   6. Divide by $a^2$ and simplify: $\dfrac{K}{6}-K-\dfrac13+\dfrac12=0 \Rightarrow -\dfrac{5K}{6}+\dfrac16=0 \Rightarrow K=\dfrac15$.

   Common Mistakes

   • Placing $M$ at $\dfrac13\overline{AB}$ measured from $B$ instead of from $A$ (the condition $\overline{AB}=3\overline{AM}$ means $M$ is $\dfrac13$ of the way from $A$ to $B$).
   • Using $\cos 60^\circ=1$ or forgetting the factor $\dfrac12$ in $\bar b.\bar c$.

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{5}$.

   ANSWER: A

6. 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 $=3^2-4(2)(5)=9-40=-31<0$.
   5. Since the discriminant is negative, there is no real value of $\lambda$ making the vectors perpendicular — the solution set is empty, $\varphi$.

   Common Mistakes

   • Forgetting to check the discriminant and assuming real roots always exist.
   • Sign slip while forming the dot product (e.g. writing $-3\lambda$ instead of $+3\lambda$).

   ✓Final answer

   The correct option is (D) — $\varphi$ (empty set).

   ANSWER: D

7. 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: $\sqrt{4^2+5^2+(-9)^2}=\sqrt{16+25+81}=\sqrt{122}$.
   6. Scale to magnitude $2$: required vector $=\dfrac{2}{\sqrt{122}}(4,5,-9)$. Since $122=2\times61$, this simplifies to $\dfrac{2}{\sqrt{2}\sqrt{61}}(4,5,-9)=\dfrac{\sqrt2}{\sqrt{61}}(4\bar i+5\bar j-9\bar k)$.

   Common Mistakes

   • Forgetting that "lying in the plane of $\bar a,\bar b$" means ANY linear combination of $\bar a,\bar b$, not just $\bar a\times\bar b$ or some fixed vector.
   • Sign/arithmetic slip in the dot products $\bar a\cdot\bar c$ or $\bar b\cdot\bar c$, which flips the ratio $\alpha:\beta$.

   ✓Final answer

   The correct option is (A) — $\dfrac{\sqrt{2}}{\sqrt{61}}(4\bar{i}+5\bar{j}-9\bar{k})$.

   ANSWER: A

8. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The plane passing through $(2,1,-3)$ and perpendicular to $3i-j+2k$ contains the points (A) $(1,5,1)$ & $(3,0,-5)$ (B) $\left(\dfrac{1}{3},3,\dfrac{1}{2}\right)$ & $\left(1,5,\dfrac{1}{2}\right)$ (C) $(3,1,-5)$ & $\left(\dfrac{1}{3},3,\dfrac{1}{2}\right)$ (D) $(1,5,3)$ & $(3,0,1)$

   ›Reveal solutionSolution

   This tests writing a plane's equation from a point and normal vector, then checking which pair of points satisfies it. The answer is option (B).

   Concept and Intuition

   A plane through point $P_0=(x_0,y_0,z_0)$ perpendicular to $\vec n=(a,b,c)$ has equation $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Once we have this equation, a point "lies in the plane" exactly when it satisfies the equation — we just substitute each candidate point and check.

   Step-by-Step Solution

   1. Normal vector $\vec n=3\hat i-\hat j+2\hat k=(3,-1,2)$, point $(2,1,-3)$.
   2. Plane equation: $3(x-2)-1(y-1)+2(z+3)=0 \Rightarrow 3x-6-y+1+2z+6=0 \Rightarrow 3x-y+2z+1=0$.
   3. Test option (A): $(1,5,1)$: $3(1)-5+2(1)+1=3-5+2+1=1\ne0$ — fails.
   4. Test option (B): $\left(\frac13,3,\frac12\right)$: $3\left(\frac13\right)-3+2\left(\frac12\right)+1=1-3+1+1=0$ — satisfies. $\left(1,5,\frac12\right)$: $3(1)-5+2\left(\frac12\right)+1=3-5+1+1=0$ — satisfies. Both points lie in the plane.
   5. Test option (C): $(3,1,-5)$: $9-1-10+1=-1\ne0$ — fails.
   6. Test option (D): $(1,5,3)$: $3-5+6+1=5\ne0$ — fails.
   7. So only option (B) has both its points satisfying the plane equation.

   Common Mistakes

   • Using the wrong sign convention when expanding the plane equation.
   • Checking only one of the two points in an option and stopping early.

   ✓Final answer

   The correct option is (B) — $\left(\dfrac{1}{3},3,\dfrac{1}{2}\right)$ & $\left(1,5,\dfrac{1}{2}\right)$.

   ANSWER: B

9. 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: $\bar d = (3t,-5t,4t) = t(3,-5,4)$.

   5. Apply $|\bar d|=\sqrt2$: $|\bar d|^2 = t^2(9+25+16)=50t^2=2 \implies t^2=\tfrac1{25}\implies t=\pm\tfrac15$.

   6. So $\bar d = \pm\dfrac15(3\bar i - 5\bar j + 4\bar k)$.

   Common Mistakes

   • Forgetting that "coplanar with $\bar a,\bar b$" through the origin means $\bar d$ is a linear combination of $\bar a,\bar b$ only — not involving $\bar c$ at all (that's reserved for the perpendicularity condition).
   • Dropping the $\pm$ sign — both directions satisfy every condition given, so both must appear in the final vector.

   ✓Final answer

   The correct option is (A) — $\pm\dfrac15(3\bar i-5\bar j+4\bar k)$.

   ANSWER: A

10. 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 $w$.
    4. So exactly the statement "$u$ is perpendicular to $v$ but not $w$" is true.

    Common Mistakes

    • Sign errors when multiplying negative components.
    • Checking only one dot product and assuming the others follow without verifying.

    ✓Final answer

    The correct option is (A) — $u$ is perpendicular to $v$ but not $w$.

    ANSWER: A

11. 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 $|\vec a|=2$, $|\vec b|=3$: $4 = 9t^2 \Rightarrow t^2=\dfrac49$.
    4. $t=\pm\dfrac23$; since $t$ is given to be a positive scalar, $t=\dfrac23$.

    Common Mistakes

    • Reporting $t=\pm2/3$ (option A) while ignoring the problem's explicit condition that $t$ is positive.
    • Forgetting that $\vec a\cdot\vec b$ terms cancel, and instead trying to solve without knowing the angle between $\vec a,\vec b$ (which isn't needed here).

    ✓Final answer

    The correct option is (C) — $t=\dfrac{2}{3}$.

    ANSWER: C

12. 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: $l-m=1$.
    7. Since $\bar c$ is a unit vector expressed in the orthonormal basis $\{\bar a,\bar b,\bar a\times\bar b\}$: $|\bar c|^2=l^2+m^2+n^2=1$, so $n^2=1-l^2-m^2$.
    8. Substitute $l=m+1$: $n^2=1-(m+1)^2-m^2=1-(m^2+2m+1)-m^2=-2m^2-2m=-2m(m+1)$.
    9. Since $l=m+1$, this is $n^2=-2m\cdot l=-2lm$.

    Common Mistakes

    • Forgetting that $\bar a.(\bar a\times\bar b)=0$ and $\bar b.(\bar a\times\bar b)=0$ (scalar triple product with a repeated vector vanishes) — this is what makes the decomposition's coefficients pop out so cleanly as dot products.
    • Not using $|\bar c|=1$ (unit vector) to get the Pythagorean relation $l^2+m^2+n^2=1$.

    ✓Final answer

    The correct option is (B) — $-2lm$.

    ANSWER: B