Q.Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $\overline{f} = i + j + k$ and $\overline{g} = 2i - j + 3k$ then the projection vector of $\overline{f}$ on $\overline{g}$ is (A) $\dfrac{2}{7}(i+j+k)$ (B) $\dfrac{2}{7}(2i-j+3k)$ (C) $\dfrac{1}{3}(i+j+k)$ (D) $\dfrac{1}{14}(2i-j+3k)$

   ›Reveal solutionSolution

   This tests the formula for the projection vector (not just scalar projection) of one vector onto another. Answer: $\dfrac27(2i-j+3k)$.

   Concept and Intuition

   The projection vector of $\overline f$ along $\overline g$ is the component of $\overline f$ that lies along $\overline g$'s direction, given by $\left(\dfrac{\overline f\cdot\overline g}{|\overline g|^2}\right)\overline g$ — the scalar projection times the unit vector along $\overline g$, written compactly using $|\overline g|^2$ in the denominator.

   Step-by-Step Solution

   1. $\overline f=i+j+k$, $\overline g=2i-j+3k$.
   2. $\overline f\cdot\overline g=(1)(2)+(1)(-1)+(1)(3)=2-1+3=4$.
   3. $|\overline g|^2=2^2+(-1)^2+3^2=4+1+9=14$.
   4. Projection vector $=\dfrac{\overline f\cdot\overline g}{|\overline g|^2}\overline g=\dfrac{4}{14}\overline g=\dfrac{2}{7}\overline g$.
   5. $=\dfrac{2}{7}(2i-j+3k)$.

   Common Mistakes

   • Dividing by $|\overline g|$ instead of $|\overline g|^2$ (that gives the scalar projection, not the vector, unless then multiplied by the unit vector).
   • Sign error in the dot product from the $-1$ component.

   ✓Final answer

   The correct option is (B) — $\dfrac{2}{7}(2i-j+3k)$.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $\bar{a} = \bar{i} - \bar{j} + 3\bar{k}$ and $\bar{b} = 3\bar{i} - 5\bar{j} + 6\bar{k}$, then the magnitude of the projection of $2\bar{a} - \bar{b}$ on $\bar{a} + \bar{b}$ is (A) $\dfrac{11\sqrt{2}}{\sqrt{10}}$ (B) $\dfrac{22}{\sqrt{10}}$ (C) $\dfrac{22}{\sqrt{133}}$ (D) $\dfrac{22}{\sqrt{5}}$

   ›Reveal solutionSolution

   This tests the projection-of-a-vector formula. Compute $2\bar a-\bar b$ and $\bar a+\bar b$, then use $\text{proj} = \dfrac{|\vec u\cdot\vec v|}{|\vec v|}$. The answer is (C).

   Concept and Intuition

   The (scalar) magnitude of the projection of $\vec u$ onto $\vec v$ measures how much of $\vec u$ lies along the direction of $\vec v$. It is given by

   $\left|\text{proj}_{\vec v}\vec u\right| = \frac{|\vec u\cdot \vec v|}{|\vec v|}.$

   This comes directly from $\vec u\cdot\vec v = |\vec u||\vec v|\cos\theta$, and $|\vec u|\cos\theta$ is exactly the signed length of the projection.

   Step-by-Step Solution

   1. Given $\bar a=\bar i-\bar j+3\bar k=(1,-1,3)$ and $\bar b=3\bar i-5\bar j+6\bar k=(3,-5,6)$.
   2. Compute $2\bar a-\bar b = (2-3,\,-2+5,\,6-6) = (-1,3,0)$.
   3. Compute $\bar a+\bar b = (1+3,\,-1-5,\,3+6) = (4,-6,9)$.
   4. Dot product: $(2\bar a-\bar b)\cdot(\bar a+\bar b) = (-1)(4)+(3)(-6)+(0)(9) = -4-18+0=-22$.
   5. Magnitude of $\bar a+\bar b$: $|\bar a+\bar b| = \sqrt{4^2+(-6)^2+9^2}=\sqrt{16+36+81}=\sqrt{133}$.
   6. Magnitude of projection $=\dfrac{|-22|}{\sqrt{133}}=\dfrac{22}{\sqrt{133}}$.

   Common Mistakes

   • Forgetting to take the absolute value of the dot product (projection magnitude is always non-negative).
   • Mixing up which vector goes in the denominator — you divide by the magnitude of the vector you are projecting onto.

   ✓Final answer

   The correct option is (C) — $\dfrac{22}{\sqrt{133}}$.

   ANSWER: C

3. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.The orthogonal projection vector of $\bar{a} = 2\bar{i} + 3\bar{j} + 3\bar{k}$ on $\bar{b} = \bar{i} - 2\bar{j} + \bar{k}$ is (A) $-\dfrac{1}{6}(2\bar{i} + 3\bar{j} + 3\bar{k})$ (B) $\dfrac{1}{6}(-\bar{i} + 2\bar{j} - \bar{k})$ (C) $\bar{i} - 2\bar{j} + \bar{k}$ (D) $-\bar{i} + 2\bar{j} - \bar{k}$

   ›Reveal solutionSolution

   Using the standard vector-projection formula $\text{proj}_{\bar b}\bar a=\dfrac{\bar a\cdot\bar b}{|\bar b|^2}\bar b$ gives $\dfrac16(-\bar i+2\bar j-\bar k)$.

   Concept and Intuition

   The orthogonal projection of $\bar a$ onto $\bar b$ is the vector component of $\bar a$ that lies along $\bar b$; it is computed by scaling $\bar b$ by the ratio $\dfrac{\bar a\cdot\bar b}{|\bar b|^2}$ (the scalar projection divided by $|\bar b|$, then re-multiplied by the unit vector along $\bar b$).

   Step-by-Step Solution

   1. $\bar a\cdot\bar b=(2)(1)+(3)(-2)+(3)(1)=2-6+3=-1$.
   2. $|\bar b|^2=1^2+(-2)^2+1^2=1+4+1=6$.
   3. Projection vector $=\dfrac{\bar a\cdot\bar b}{|\bar b|^2}\bar b=\dfrac{-1}{6}(\bar i-2\bar j+\bar k)$.
   4. Distribute the $-\dfrac16$: $=-\dfrac16\bar i+\dfrac26\bar j-\dfrac16\bar k=\dfrac16(-\bar i+2\bar j-\bar k)$.

   Common Mistakes

   • Forgetting to square $|\bar b|$ in the denominator (using $|\bar b|$ instead of $|\bar b|^2$ gives the scalar projection's magnitude scaling wrong).
   • Sign error carrying the negative dot product through the distribution.

   ✓Final answer

   The correct option is (B) — $\dfrac16(-\bar i+2\bar j-\bar k)$.

   ANSWER: B

4. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.Given $\vec{a} = 3\hat{i} - \hat{j}$, $\vec{b} = 2\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{b} = \vec{b}_1 + \vec{b}_2$ where $\vec{b}_1$ is parallel to $\vec{a}$ and $\vec{b}_2$ is perpendicular to $\vec{a}$ then $\vec{b}_2$ is equal to (A) $\dfrac{1}{2}\hat{i} + \dfrac{3}{2}\hat{j} - 3\hat{k}$ (B) $\dfrac{1}{2}\hat{i} - \dfrac{3}{2}\hat{j} + 3\hat{k}$ (C) $\dfrac{1}{2}\hat{i} + \dfrac{3}{2}\hat{j} + 3\hat{k}$ (D) $\dfrac{1}{2}\hat{i} - \dfrac{3}{2}\hat{j} - 3\hat{k}$

   ›Reveal solutionSolution

   $\vec b_2$ is $\vec b$ minus its projection onto $\vec a$; computing that projection gives $\vec b_2=\frac12\hat i+\frac32\hat j-3\hat k$.

   Concept and Intuition

   Any vector $\vec b$ can be decomposed into a component parallel to a given direction $\vec a$ (the vector projection) and a component perpendicular to it — the perpendicular part is just what's left after subtracting the parallel part.

   Step-by-Step Solution

   1. The parallel component is $\vec b_1 = \dfrac{\vec a\cdot\vec b}{|\vec a|^2}\,\vec a$.
   2. $\vec a=3\hat i-\hat j$, $\vec b = 2\hat i+\hat j-3\hat k$. Compute $\vec a\cdot\vec b = 3(2)+(-1)(1)+0(-3) = 6-1+0=5$.
   3. $|\vec a|^2 = 3^2+(-1)^2 = 9+1=10$.
   4. So $\vec b_1 = \dfrac{5}{10}(3\hat i-\hat j) = \dfrac12(3\hat i-\hat j) = \dfrac32\hat i-\dfrac12\hat j$.
   5. $\vec b_2 = \vec b-\vec b_1 = \left(2\hat i+\hat j-3\hat k\right) - \left(\dfrac32\hat i-\dfrac12\hat j\right) = \dfrac12\hat i+\dfrac32\hat j-3\hat k$.
   6. Quick check: $\vec b_2\cdot\vec a = \dfrac12(3)+\dfrac32(-1)+(-3)(0) = 1.5-1.5=0$ ✓, confirming perpendicularity.

   Common Mistakes

   • Sign slip subtracting $\vec b_1$'s $\hat j$-component (it's $-\frac12\hat j$, so subtracting it adds $+\frac12\hat j$ to $\vec b$'s own $+1\hat j$).
   • Forgetting to carry the untouched $-3\hat k$ term from $\vec b$ into $\vec b_2$ (since $\vec a$ has no $\hat k$ component).

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{2}\hat{i} + \dfrac{3}{2}\hat{j} - 3\hat{k}$.

   ANSWER: A

5. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If $\bar a = 4\bar i + 6\bar j$, $\bar b = 3\bar j + 4\bar k$ and $\bar c$ is the projection vector of $\bar a$ on $\bar b$, then $\bar c$ and $|\bar c|$ respectively are (A) $\dfrac{18}{25}\bar b, \dfrac{18}{5}$ (B) $\dfrac{18}{5}\bar b, 18$ (C) $\dfrac{25}{18}\bar b, \dfrac{18}{5}$ (D) $\dfrac{5}{18}\bar b, \dfrac{5}{18}$

   ›Reveal solutionSolution

   The projection vector formula $\bar c=\dfrac{\bar a\cdot\bar b}{|\bar b|^2}\bar b$ gives both $\bar c$ and its magnitude directly.

   Concept and Intuition

   The projection (vector component) of $\bar a$ along $\bar b$ is the vector $\bar c$ along $\bar b$'s direction whose length is $\bar a$'s component along $\bar b$. The formula packages both the direction (a scalar multiple of $\bar b$) and the magnitude in one expression.

   Step-by-Step Solution

   1. $\bar a = 4\bar i+6\bar j+0\bar k$, $\bar b = 0\bar i+3\bar j+4\bar k$.
   2. $\bar a\cdot\bar b = 4(0)+6(3)+0(4) = 18$.
   3. $|\bar b|^2 = 0^2+3^2+4^2 = 25$, so $|\bar b|=5$.
   4. Projection vector: $\bar c = \dfrac{18}{25}\bar b$.
   5. Magnitude: $|\bar c| = \dfrac{18}{25}\times 5 = \dfrac{18}{5}$.

   Common Mistakes

   • Confusing the projection vector (a multiple of $\bar b$) with the projection scalar $\bar a\cdot\bar b/|\bar b| = 18/5$ — here both happen to share the number $18/5$, which is a coincidence worth double-checking, not a shortcut to rely on generally.

   ✓Final answer

   The correct option is (A) — $\dfrac{18}{25}\bar b,\ \dfrac{18}{5}$.

   ANSWER: A

6. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Let $\bar{a} = 2\bar{i} + 2\bar{j} - \bar{k}$, $\bar{b} = \bar{i} - 2\bar{j} + \bar{k}$ be two vectors. If $\bar{l}$ is the component vector of $\bar{b}$ parallel to $\bar{a}$ and $\bar{m}$ is the component vector of $\bar{a}$ perpendicular to $\bar{b}$, then $3\bar{l} + 2\bar{m} =$ (A) $\bar{i} - 2\bar{j} + 2\bar{k}$ (B) $\bar{i} + 3\bar{j}$ (C) $3\bar{i}$ (D) $-\bar{j} + 2\bar{k}$

   ›Reveal solutionSolution

   Compute the vector projection $\bar l$ of $\bar b$ onto $\bar a$ and the perpendicular component $\bar m$ of $\bar a$ relative to $\bar b$, then combine linearly. Answer: $3\bar i$.

   Concept and Intuition

   The component of $\bar b$ parallel to $\bar a$ is the vector projection $\bar l=\dfrac{\bar a\cdot\bar b}{|\bar a|^2}\bar a$. The component of $\bar a$ perpendicular to $\bar b$ is what's left after removing $\bar a$'s projection onto $\bar b$: $\bar m=\bar a-\dfrac{\bar a\cdot\bar b}{|\bar b|^2}\bar b$.

   Step-by-Step Solution

   1. $\bar a=(2,2,-1)$, $\bar b=(1,-2,1)$. $\bar a\cdot\bar b=2(1)+2(-2)+(-1)(1)=2-4-1=-3$.
   2. $|\bar a|^2=4+4+1=9$. So $\bar l=\dfrac{-3}{9}\bar a=-\dfrac13(2,2,-1)=\left(-\dfrac23,-\dfrac23,\dfrac13\right)$.
   3. $|\bar b|^2=1+4+1=6$. So $\dfrac{\bar a\cdot\bar b}{|\bar b|^2}\bar b=\dfrac{-3}{6}(1,-2,1)=\left(-\dfrac12,1,-\dfrac12\right)$.
   4. $\bar m=\bar a-\left(-\dfrac12,1,-\dfrac12\right)=\left(2+\dfrac12,\ 2-1,\ -1+\dfrac12\right)=\left(\dfrac52,1,-\dfrac12\right)$.
   5. $3\bar l=3\left(-\dfrac23,-\dfrac23,\dfrac13\right)=(-2,-2,1)$.
   6. $2\bar m=2\left(\dfrac52,1,-\dfrac12\right)=(5,2,-1)$.
   7. $3\bar l+2\bar m=(-2+5,\ -2+2,\ 1-1)=(3,0,0)=3\bar i$.

   Common Mistakes

   • Confusing the two formulas — computing $\bar m$ as if it were the perpendicular component of $\bar b$ w.r.t. $\bar a$ instead of $\bar a$ w.r.t. $\bar b$ (the problem explicitly swaps the roles).
   • Sign slips in the subtraction step when forming $\bar m$.

   ✓Final answer

   The correct option is (C) — $3\bar{i}$.

   ANSWER: C

7. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.Let $\bar{a}=4\bar{i}+3\bar{j}$ and $\bar{b}$ be two perpendicular vectors in the XOY-plane. A vector $\bar{c}$ in the same plane and having projections 1 and 2 respectively on $\bar{a}$ and $\bar{b}$ is (A) $\bar{i}+2\bar{j}$ (B) $2\bar{i}+\bar{j}$ (C) $\bar{i}-2\bar{j}$ (D) $2\bar{i}-\bar{j}$

   ›Reveal solutionSolution

   Uses that two perpendicular unit vectors form an orthonormal basis of the plane, so $\bar c$ is rebuilt directly from its two given projections; the answer is (D).

   Concept and Intuition

   If $\hat a,\hat b$ are two perpendicular unit vectors spanning a plane, any vector $\bar c$ in that plane can be written as $\bar c=(\bar c\cdot\hat a)\hat a+(\bar c\cdot\hat b)\hat b$ — the coefficients are exactly the (scalar) projections of $\bar c$ onto each axis, because $\hat a,\hat b$ act like the $x,y$ axes rotated into place.

   Step-by-Step Solution

   1. $|\bar a|=\sqrt{4^2+3^2}=5$, so $\hat a=\dfrac{1}{5}(4,3)$.
   2. $\bar b\perp\bar a$ in the plane, so its unit vector is $\hat b=\dfrac{1}{5}(3,-4)$ (rotate $\hat a$ by $90^\circ$; the other perpendicular choice just swaps signs and is ruled out below by matching an option).
   3. Given projections: $\bar c\cdot\hat a=1$ and $\bar c\cdot\hat b=2$, so $\bar c=1\cdot\hat a+2\cdot\hat b=\dfrac{1}{5}(4,3)+\dfrac{2}{5}(3,-4)=\dfrac{1}{5}(4+6,\,3-8)=\dfrac{1}{5}(10,-5)=(2,-1)$.
   4. So $\bar c=2\bar i-\bar j$.
   5. Check: $\bar c\cdot\bar a=(2)(4)+(-1)(3)=5$, projection on $\bar a$ $=5/5=1$ ✓. $\bar c\cdot(3,-4)=6+4=10$, projection on $\bar b=3\bar i-4\bar j$ (magnitude 5) is $10/5=2$ ✓.

   Common Mistakes

   • Picking the wrong sign/direction for $\hat b$ without checking against the projections — always verify the final vector's projections numerically.
   • Forgetting to divide by $|\bar a|,|\bar b|$ when "projection" means the scalar (not vector) component.

   ✓Final answer

   The correct option is (D) — $2\bar{i}-\bar{j}$.

   ANSWER: D

8. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Let $\bar{a} = 4\bar{i} + 3\bar{j}$ and $\bar{b}$ be two vectors in XOY plane and let $\bar{a}$ be perpendicular to $\bar{b}$. Then a vector $\bar{c}$ in the same plane and having projections 1 and 2 respectively on $\bar{a}$ and $\bar{b}$ is (A) $\bar{i} + 2\bar{j}$ (B) $2\bar{i} + \bar{j}$ (C) $\bar{i} - 2\bar{j}$ (D) $2\bar{i} - \bar{j}$

   ›Reveal solutionSolution

   Resolving $\bar c$ along the perpendicular directions of $\bar a$ and $\bar b$ using the given scalar projections 1 and 2 yields $\bar c = 2\bar i - \bar j$.

   Concept and Intuition

   Since $\bar a$ and $\bar b$ are perpendicular vectors in the plane, their unit vectors form an orthonormal basis for that plane. Any vector $\bar c$ in the plane can be written as (projection on $\bar a$)$\times$(unit vector along $\bar a$) + (projection on $\bar b$)$\times$(unit vector along $\bar b$).

   Step-by-Step Solution

   1. $|\bar a|=\sqrt{16+9}=5$, so unit vector along $\bar a$ is $\hat a=(4/5,3/5)$.
   2. A unit vector perpendicular to $\hat a$ in the plane is $\hat b=(3/5,-4/5)$ (the other perpendicular choice is $(-3/5,4/5)$; we pick the sign that is consistent with the answer, as is standard when $\bar b$'s orientation isn't otherwise pinned down).
   3. $\bar c = 1\cdot\hat a + 2\cdot\hat b = \left(\dfrac45+\dfrac65,\ \dfrac35-\dfrac85\right) = (2,-1)$.
   4. Verify: projection of $\bar c$ on $\bar a$ is $\bar c\cdot\hat a = 2\cdot\tfrac45+(-1)\cdot\tfrac35=\tfrac85-\tfrac35=1$ ✓. Projection on $\bar b$ is $\bar c\cdot\hat b=2\cdot\tfrac35+(-1)\cdot(-\tfrac45)=\tfrac65+\tfrac45=2$ ✓.
   5. So $\bar c = 2\bar i-\bar j$.

   Common Mistakes

   • Using the wrong perpendicular direction sign, which flips the answer to $\bar i-2\bar j$-type forms that don't check out.
   • Forgetting to verify the reconstructed vector actually reproduces both stated projections.

   ✓Final answer

   The correct option is (D) — $2\bar{i} - \bar{j}$.

   ANSWER: D

9. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $\vec a=\vec i+3\vec j+13\vec k$ and $\vec b=2\vec i-4\vec j+3\vec k$ are two vectors, then the component vector of $\vec a$ perpendicular to $\vec b$ is (A) $\vec i-\vec j-2\vec k$ (B) $3\vec i+3\vec j+2\vec k$ (C) $-\vec i+7\vec j+10\vec k$ (D) $4\vec i+5\vec j+4\vec k$

   ›Reveal solutionSolution

   Since $\vec a\cdot\vec b=|\vec b|^2=29$, the projection of $\vec a$ onto $\vec b$ is simply $\vec b$ itself, so the perpendicular component is $\vec a-\vec b=-\vec i+7\vec j+10\vec k$.

   Concept and Intuition

   Any vector $\vec a$ splits uniquely into a component parallel to $\vec b$ (the projection) and a component perpendicular to $\vec b$: $\vec a=\vec a_{\parallel}+\vec a_{\perp}$, where $\vec a_{\parallel}=\dfrac{\vec a\cdot\vec b}{|\vec b|^2}\vec b$. A nice numerical coincidence here ($\vec a\cdot\vec b=|\vec b|^2$) makes the projection scalar exactly $1$, simplifying the arithmetic.

   Step-by-Step Solution

   1. $\vec a=(1,3,13)$, $\vec b=(2,-4,3)$.
   2. $\vec a\cdot\vec b=1(2)+3(-4)+13(3)=2-12+39=29$.
   3. $|\vec b|^2=2^2+(-4)^2+3^2=4+16+9=29$.
   4. Projection scalar $=\dfrac{\vec a\cdot\vec b}{|\vec b|^2}=\dfrac{29}{29}=1$, so the parallel component is $1\cdot\vec b=\vec b=(2,-4,3)$.
   5. Perpendicular component $=\vec a-\vec a_{\parallel}=(1-2,\ 3-(-4),\ 13-3)=(-1,7,10)=-\vec i+7\vec j+10\vec k$.

   Common Mistakes

   • Forgetting to subtract the parallel component from $\vec a$ (some might mistakenly report the parallel component itself as the answer).
   • Sign error in $3-(-4)=7$.

   ✓Final answer

   The correct option is (C) — $-\vec i+7\vec j+10\vec k$.

   ANSWER: C

10. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.Let $\overline{a} = 3\overline{i} + 4\overline{j} - 5\overline{k}, \overline{b} = 2\overline{i} + \overline{j} - 2\overline{k}$. The projection of the sum of the vectors $\overline{a}, \overline{b}$ on the vector perpendicular to the plane of $\overline{a}, \overline{b}$ is (A) $0$ (B) $4\sqrt{2}$ (C) $7\sqrt{2}$ (D) $\dfrac{1}{\sqrt{2}}$

    ›Reveal solutionSolution

    This tests the basic fact that the cross product of two vectors is perpendicular to every vector lying in their span. Answer: $0$.

    Concept and Intuition

    "The vector perpendicular to the plane of $\overline a,\overline b$" is (a scalar multiple of) $\overline a\times\overline b$. By definition, $\overline a\times\overline b$ is orthogonal to both $\overline a$ and $\overline b$ — and hence orthogonal to every vector that lies in the plane they span, including $\overline a+\overline b$. A projection of a vector onto something perpendicular to it is always $0$.

    Step-by-Step Solution

    1. Let $\overline n=\overline a\times\overline b$, the vector perpendicular to the plane containing $\overline a$ and $\overline b$.
    2. Any vector $\overline v$ that can be written as $\lambda\overline a+\mu\overline b$ lies in this plane, so $\overline v\cdot\overline n=0$.
    3. $\overline a+\overline b$ is exactly such a combination (with $\lambda=\mu=1$), so $(\overline a+\overline b)\cdot\overline n=0$.
    4. The projection of $\overline a+\overline b$ on $\overline n$ is $\dfrac{(\overline a+\overline b)\cdot\overline n}{|\overline n|}=0$.
    5. No coordinate computation of $\overline a,\overline b$ is even needed — the result follows purely from the geometry.

    Common Mistakes

    • Wasting time computing $\overline a\times\overline b$ explicitly and then the projection — the answer is immediate from perpendicularity.
    • Confusing "projection onto the normal" with "projection onto $\overline a$ or $\overline b$" (which would not be zero).

    ✓Final answer

    The correct option is (A) — $0$.

    ANSWER: A

11. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.$3\bar i+\bar j+\bar k$, $2\bar i+\bar k$, $\bar i+5\bar j$ are the position vectors of three non collinear points A, B, C respectively. If the perpendicular drawn from C onto $\overline{AB}$ meets $\overline{AB}$ at the point $a\bar i+b\bar j+c\bar k$, then $a+b+c=$ (A) 5 (B) 3 (C) 7 (D) 9

    ›Reveal solutionSolution

    Find the foot of the perpendicular from $C$ onto line $AB$ using the perpendicularity condition; it comes out to $(4,2,1)$, so $a+b+c=7$.

    Concept and Intuition

    The foot of the perpendicular from an external point onto a line is the point on the line whose connecting vector to the external point is orthogonal to the line's direction vector — this converts a geometry problem into one linear (dot-product) equation in the parameter $t$.

    Step-by-Step Solution

    1. $A=(3,1,1)$, $B=(2,0,1)$, $C=(1,5,0)$ (from the given position vectors).
    2. $\vec{AB}=B-A=(-1,-1,0)$.
    3. General point on line $AB$: $P=A+t\,\vec{AB}=(3-t,\,1-t,\,1)$.
    4. $\vec{CP}=P-C=(2-t,\,-4-t,\,1)$.
    5. Perpendicularity: $\vec{CP}\cdot\vec{AB}=0 \Rightarrow -(2-t)-(-4-t)=0 \Rightarrow 2t+2=0 \Rightarrow t=-1$.
    6. $P=(3-(-1),\,1-(-1),\,1)=(4,2,1)$, so $a=4,b=2,c=1$ and $a+b+c=7$.

    Common Mistakes

    • Forgetting the $z$-coordinate stays fixed at $1$ throughout since both $A$ and $B$ have $z=1$.
    • Sign errors when expanding the dot product with the negative direction components.

    ✓Final answer

    The correct option is (C) — $7$.

    ANSWER: C

12. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.Let $\bar{a}=2\bar{i}+\bar{j}+3\bar{k}$, $\bar{b}=3\bar{i}+3\bar{j}+\bar{k}$ and $\bar{c}=\bar{i}-2\bar{j}+3\bar{k}$ be three vectors. If $\bar{r}$ is a vector such that $\bar{r}\times\bar{a}=\bar{r}\times\bar{b}$ and $\bar{r}.\bar{c}=18$, then the magnitude of the orthogonal projection of $4\bar{i}+3\bar{j}-\bar{k}$ on $\bar{r}$ is (A) $4$ (B) $6$ (C) $12$ (D) $24$

    ›Reveal solutionSolution

    This tests using $\bar r\times\bar a=\bar r\times\bar b$ to pin down the direction of $\bar r$, a dot-product condition to fix its magnitude, and then computing a scalar projection. The projection magnitude is $4$.

    Concept and Intuition

    If $\bar r\times\bar a=\bar r\times\bar b$, then $\bar r\times(\bar a-\bar b)=\bar 0$, which forces $\bar r$ to be parallel to $\bar a-\bar b$ (assuming $\bar r\ne\bar 0$ and $\bar a\ne\bar b$). Once the direction of $\bar r$ is known, a single scalar condition like $\bar r\cdot\bar c=18$ fixes the scaling factor completely, after which any projection is a routine dot-product computation.

    Step-by-Step Solution

    1. $\bar a-\bar b=(2-3,1-3,3-1)=(-1,-2,2)$.
    2. Since $\bar r\times\bar a=\bar r\times\bar b\Rightarrow\bar r\times(\bar a-\bar b)=\bar0$, $\bar r$ is parallel to $(-1,-2,2)$: write $\bar r=t(-1,-2,2)$.
    3. Use $\bar r\cdot\bar c=18$ with $\bar c=(1,-2,3)$: $t[(-1)(1)+(-2)(-2)+(2)(3)]=t(-1+4+6)=9t=18\Rightarrow t=2$.
    4. So $\bar r=(-2,-4,4)$, and $|\bar r|=\sqrt{4+16+16}=\sqrt{36}=6$.
    5. Magnitude of the orthogonal projection of $\bar v=(4,3,-1)$ on $\bar r$ is $\dfrac{|\bar v\cdot\bar r|}{|\bar r|}$.
    6. $\bar v\cdot\bar r=4(-2)+3(-4)+(-1)(4)=-8-12-4=-24$.
    7. Magnitude $=\dfrac{|-24|}{6}=4$.

    Common Mistakes

    • Forgetting the absolute value when computing projection magnitude (sign of the dot product doesn't matter for magnitude).
    • Sign slip when computing $\bar a-\bar b$ vs $\bar b-\bar a$ — either direction works since $\bar r$ is only fixed up to sign by the cross-product condition, but the magnitude of $t$ then adjusts consistently.

    ✓Final answer

    The correct option is (A) — $4$.

    ANSWER: A