Q.Find the projection of the vector $\hat{i}+3\hat{j}+7\hat{k}$ on the vector $7\hat{i}-\hat{j}+8\hat{k}$.

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. $|\overline g|^2=2^2+(-1)^2+3^2=4+1+9=14$. …

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2. 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 $=\dfrac{\bar a\cdot\bar b}{|\bar b|^2}\bar b=\dfrac{-1}{6}(\bar i-2\bar j+\bar k)$. …

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3. 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: $(2\bar a-\bar b)\cdot(\bar a+\bar b) = (-1)(4)+(3)(-6)+(0)(9) = -4-18+0=-22$. …

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4. 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 $(\overline a+\overline b)\cdot\overline n=0$. …

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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 …

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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. $\bar m=\bar a-\left(-\dfrac12,1,-\dfrac12\right)=\left(2+\dfrac12,\ 2-1,\ -1+\dfrac12\right)=\left(\dfrac52,1,-\dfrac12\right)$. …

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7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.Let $\bar{a} \times \bar{b} = 7\bar{i} - 5\bar{j} - 4\bar{k}$ and $\bar{a} = \bar{i} + 3\bar{j} - 2\bar{k}$. If the length of projection of $\bar{b}$ on $\bar{a}$ is $\dfrac{8}{\sqrt{14}}$, then $|\bar{b}| =$ (A) 121 (B) $\sqrt{12}$ (C) $\sqrt{11}$ (D) 144

   ›Reveal solutionSolution

   Combine the projection formula with the identity linking cross product magnitude, dot product, and the two vector magnitudes; solving gives $|\bar b|=\sqrt{11}$.

   Concept and Intuition

   For any two vectors, $|\bar a\times\bar b|^2+(\bar a.\bar b)^2=|\bar a|^2|\bar b|^2$ (this follows from $|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta$ and $\bar a.\bar b=|\bar a||\bar b|\cos\theta$). The projection length of $\bar b$ on $\bar a$ is $\dfrac{\bar a.\bar b}{|\bar a|}$, which directly gives us $\bar a.\bar b$.

   Step-by-Step Solution

   1. $|\bar a|^2=1^2+3^2+(-2)^2=14$.
   2. Projection of $\bar b$ on $\bar a$: $\dfrac{\bar a.\bar b}{|\bar a|}=\dfrac{8}{\sqrt{14}}\Rightarrow \bar a.\bar b = 8$.
   3. $|\bar a\times\bar b|^2 = 7^2+(-5)^2+(-4)^2 = 49+25+16=90$. …

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8. 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. $|\vec b|^2=2^2+(-4)^2+3^2=4+16+9=29$. …

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9. 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. $\vec{CP}=P-C=(2-t,\,-4-t,\,1)$. …

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10. 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. $\vec b_2 = \vec b-\vec b_1 = \left(2\hat i+\hat j-3\hat k\right) - \left(\dfrac32\hat i-\dfrac12\hat j\right) = \dfrac12\hat i+\dfrac32\hat j-3\hat k$. …

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11. 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 $\bar c=2\bar i-\bar j$. …

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12. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.If the projections of the line segment $\overline{AB}$ on $XY$, $YZ$ and $ZX$ planes are $\sqrt{15}, \sqrt{46}, 7$ respectively, then the projection of $\overline{AB}$ on $y$-axis is ______. (A) $9$ (B) $3$ (C) $4$ (D) $7$

    ›Reveal solutionSolution

    Adding the three plane-projection relations gives the total $l^2+m^2+n^2$; subtracting each given value in turn recovers each axis component squared. Answer: $3$.

    Concept and Intuition

    If $(l,m,n)$ are the components of $\vec{AB}$ along $x,y,z$, then the projection onto a coordinate plane equals the length of the vector with the third (excluded) component dropped: projection on $XY$-plane $=\sqrt{l^2+m^2}$ (drops $n$), on $YZ$-plane $=\sqrt{m^2+n^2}$ (drops $l$), on $ZX$-plane $=\sqrt{n^2+l^2}$ (drops $m$). Adding all three double-counts every component once, giving the total $l^2+m^2+n^2$ directly.

    Step-by-Step Solution

    1. $l^2+m^2=15$, $m^2+n^2=46$, $n^2+l^2=49$.
    2. Sum: $2(l^2+m^2+n^2)=15+46+49=110\Rightarrow l^2+m^2+n^2=55$.
    3. Subtracting each plane relation from the total recovers the excluded (axis) component squared: $55-15=40$, $55-46=9$, $55-49=6$. …

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