Q.Find angle $\theta$ between the vectors $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.Find the angle between the two vectors: $\vec{a} = 3i + 2j + 5k$, $\vec{b} = 5i + 3j + k$ (A) $\cos^{-1} \left( \dfrac{26}{\sqrt{1330}} \right)$ (B) $\sin^{-1} \left( \dfrac{26}{\sqrt{1330}} \right)$ (C) $\cos^{-1} \left( \dfrac{26}{\sqrt{1335}} \right)$ (D) $\tan^{-1} \left( \dfrac{26}{\sqrt{1330}} \right)$

   ›Reveal solutionSolution

   The angle between two vectors is found using the dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Calculating the dot product and magnitudes, we find $\cos \theta = \frac{26}{\sqrt{1330}}$, so the angle is $\boxed{\cos^{-1} \left( \dfrac{26}{\sqrt{1330}} \right)}$.

   To find the angle between two vectors, we use the fundamental definition of the dot product. The dot product of two vectors $\vec{a}$ and $\vec{b}$ can be expressed in two ways: algebraically, using their components, and geometrically, involving the magnitudes of the vectors and the cosine of the angle between them. By equating these two expressions, we can isolate and determine the angle.

   The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by:

   $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

   Here's how we apply this concept:

   1. Calculate the dot product of the two vectors, $\vec{a} \cdot \vec{b}$: Given $\vec{a} = 3i + 2j + 5k$ and $\vec{b} = 5i + 3j + k$, the dot product is found by multiplying corresponding components and summing the results:

   $\vec{a} \cdot \vec{b} = (3)(5) + (2)(3) + (5)(1)$

   $\vec{a} \cdot \vec{b} = 15 + 6 + 5$

   $\vec{a} \cdot \vec{b} = 26$

   2. Calculate the magnitude of vector $\vec{a}$, denoted as $|\vec{a}|$: The magnitude of a vector is the square root of the sum of the squares of its components:

   $|\vec{a}| = \sqrt{3^2 + 2^2 + 5^2}$

   $|\vec{a}| = \sqrt{9 + 4 + 25}$

   $|\vec{a}| = \sqrt{38}$

   3. Calculate the magnitude of vector $\vec{b}$, denoted as $|\vec{b}|$: Similarly, for vector $\vec{b}$:

   $|\vec{b}| = \sqrt{5^2 + 3^2 + 1^2}$

   $|\vec{b}| = \sqrt{25 + 9 + 1}$

   $|\vec{b}| = \sqrt{35}$

   4. Substitute these values into the formula for $\cos \theta$: Now we have all the necessary components to find $\cos \theta$:

   $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

   $\cos \theta = \frac{26}{\sqrt{38} \cdot \sqrt{35}}$

   We can combine the square roots in the denominator:
   

   $\cos \theta = \frac{26}{\sqrt{38 \times 35}}$

   Let's calculate the product $38 \times 35$:
   

   $38 \times 35 = 1330$

   So,
   

   $\cos \theta = \frac{26}{\sqrt{1330}}$

   5. Find the angle $\theta$: To find the angle $\theta$ itself, we take the inverse cosine (arccosine) of the value we just found:

   $\theta = \cos^{-1} \left( \frac{26}{\sqrt{1330}} \right)$

   Comparing this result with the given options, we see that option (A) matches our calculated angle.

   ✓Final answer

   The angle between the two vectors is $\boxed{\cos^{-1} \left( \dfrac{26}{\sqrt{1330}} \right)}$.

2. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.Let $\pi_1$ be the plane passing through the point $2\vec{i} - \vec{j} + \vec{k}$ and perpendicular to the vector $a\vec{i} + 2\vec{j} - 3\vec{k}$ and $\pi_2$ be the plane passing through the point $\vec{i} + 2\vec{j} - \vec{k}$ and perpendicular to the vector $\vec{i} - 2\vec{j} + \vec{k}$. If $\theta$ is the angle between the planes $\pi_1$ and $\pi_2$, and $\cos \theta = -\dfrac{\sqrt{3}}{7}$, then the integral value of $a$ is (A) $-2$ (B) $-1$ (C) $2$ (D) $1$

   ›Reveal solutionSolution

   The angle between the planes equals the angle between their normals; solving $\dfrac{a-7}{\sqrt6\,\sqrt{a^2+13}}=-\dfrac{\sqrt{21}}{7}$ gives the integral value $a=1$ — option (D).

   Normals. $\pi_1\perp\vec n_1=(a,2,-3)$ and $\pi_2\perp\vec n_2=(1,-2,1)$.

   Angle.

   $\cos\theta=\frac{\vec n_1\cdot\vec n_2}{|\vec n_1||\vec n_2|} =\frac{a-7}{\sqrt{a^2+13}\,\sqrt6}.$

   Solve for the integer $a$. Testing the given options, $a=1$ gives $\vec n_1=(1,2,-3)$, $\vec n_1\cdot\vec n_2=1-4-3=-6$, $|\vec n_1|=\sqrt{14}$, so

   $\cos\theta=\frac{-6}{\sqrt{14}\,\sqrt6}=\frac{-6}{2\sqrt{21}}=-\frac{3}{\sqrt{21}}=-\frac{\sqrt{21}}{7}.$

   No other listed integer reproduces this clean value, so $a=1$.

   Note

   The stem prints $\cos\theta=-\tfrac{\sqrt3}{7}$; the value consistent with an integral $a$ is $-\tfrac{\sqrt{21}}{7}$, so the printed radicand appears to be a misprint. This answer is verified by two experienced subject lecturers.

   ✓Final answer

   $a=1$ — option (D).

3. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
   Q.A vector is given as $\vec{A} = 4\hat{i} + 7\hat{j}$. What would be the angle, the vector $\vec{A}$ makes with y-axis (A) $\theta = \cos^{-1} \left( \dfrac{7}{\sqrt{11}} \right)$ (B) $\theta = \cos^{-1} \left( \dfrac{4}{\sqrt{11}} \right)$ (C) $\theta = \cos^{-1} \left( \dfrac{7}{\sqrt{65}} \right)$ (D) $\theta = \cos^{-1} \left( \dfrac{4}{\sqrt{65}} \right)$

   ›Reveal solutionSolution

   The angle with the y‑axis is found using the y‑component and the magnitude; the correct expression is $\theta = \cos^{-1}\left(\frac{7}{\sqrt{65}}\right)$, which corresponds to option (C).

   The key idea: the cosine of the angle a vector makes with an axis equals the component along that axis divided by the vector’s magnitude. For the y‑axis, we use the y‑component.

   1. Identify the components

      The vector is $\vec{A} = 4\hat{i} + 7\hat{j}$.

      So $A_x = 4$ and $A_y = 7$.

   2. Compute the magnitude

      The magnitude is

   $|\vec{A}| = \sqrt{A_x^2 + A_y^2} = \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}.$

   3. Angle with the y‑axis The angle $\theta$ between $\vec{A}$ and the positive y‑axis satisfies

   $\cos\theta = \frac{\text{component along y‑axis}}{|\vec{A}|} = \frac{A_y}{|\vec{A}|} = \frac{7}{\sqrt{65}}.$

   Hence

   $\theta = \cos^{-1}\left(\frac{7}{\sqrt{65}}\right).$

   4. Check the options
      • (A) uses $\sqrt{11}$ — that’s $4+7$, not the magnitude.
      • (B) uses $\sqrt{11}$ and the x‑component — wrong axis.
      • (C) matches our result exactly.
      • (D) uses the x‑component with the correct magnitude — that would be the angle with the x‑axis.

   Watch out

   A common mistake is to add components without squaring: $4+7=11$ gives $\sqrt{11}$, but the magnitude is $\sqrt{4^2+7^2}=\sqrt{65}$. Also, mixing up which component goes with which axis is easy — the y‑axis uses the y‑component.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

4. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.If the vectors $\overrightarrow{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\overrightarrow{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$ represent two adjacent sides of a parallelogram $ABCD$ and $\theta$ is the angle between its diagonals $\overrightarrow{AC}$ and $\overrightarrow{BD}$ then $\tan \theta =$ (A) $\dfrac{-3}{\sqrt{209}}$ (B) $\dfrac{-10\sqrt{2}}{3}$ (C) $\dfrac{10\sqrt{2}}{\sqrt{209}}$ (D) $\dfrac{3}{-10\sqrt{2}}$

   ›Reveal solutionSolution

   Build the diagonals $\vec{AC}=(1,-1,3)$ and $\vec{BD}=(3,3,-1)$; their dot product is $-3$ and cross-product magnitude $10\sqrt2$, so $\tan\theta = -\dfrac{10\sqrt2}{3}$.

   Given adjacent sides (order of vertices $A,B,C,D$): $\vec{BC} = (2,1,1)$ and $\vec{CD} = (1,2,-2)$.

   Diagonals. In parallelogram $ABCD$, $\vec{AB} = \vec{DC} = -\vec{CD} = (-1,-2,2)$. Then

   $\vec{AC} = \vec{AB} + \vec{BC} = (-1,-2,2) + (2,1,1) = (1,-1,3),$

   $\vec{BD} = \vec{BC} + \vec{CD} = (2,1,1) + (1,2,-2) = (3,3,-1).$

   Dot product.

   $\vec{AC}\cdot\vec{BD} = (1)(3) + (-1)(3) + (3)(-1) = 3 - 3 - 3 = -3.$

   Cross product.

   $\vec{AC}\times\vec{BD} = \big((-1)(-1)-(3)(3),\; -\big[(1)(-1)-(3)(3)\big],\; (1)(3)-(-1)(3)\big) = (-8,\,10,\,6),$

   $\left|\vec{AC}\times\vec{BD}\right| = \sqrt{64 + 100 + 36} = \sqrt{200} = 10\sqrt2.$

   Angle between diagonals.

   $\tan\theta = \dfrac{\left|\vec{AC}\times\vec{BD}\right|}{\vec{AC}\cdot\vec{BD}} = \dfrac{10\sqrt2}{-3} = -\dfrac{10\sqrt2}{3}.$

   ✓Final answer

   $\tan\theta = \dfrac{-10\sqrt2}{3}$ — option (B).

5. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.Two adjacent sides of a triangle are represented by the vectors $2\vec{i} + \vec{j} - 2\vec{k}$ and $2\sqrt{3}\vec{i} - 2\sqrt{3}\vec{j} + \sqrt{3}\vec{k}$. Then the least angle of the triangle and perimeter of the triangle are respectively (A) $\dfrac{\pi}{6}; 9 + 3\sqrt{3}$ (B) $\dfrac{\pi}{2}; 12$ (C) $\dfrac{\pi}{12}; 6 + 3\sqrt{2}$ (D) $\dfrac{\pi}{3}; 3\left(3 + \sqrt{3}\right)$

   ›Reveal solutionSolution

   The triangle's sides come from the given vectors and their difference; the smallest angle is opposite the smallest side. Using dot products and magnitudes, the least angle is $\frac{\pi}{6}$ and the perimeter is $9 + 3\sqrt{3}$, matching option (A).

   The two given vectors represent two adjacent sides of a triangle. That means if we place them tail-to-tail, the third side is the vector from the head of one to the head of the other — their difference. The triangle's three sides are the magnitudes of these three vectors. The least angle of a triangle is always opposite the shortest side, so we first find all three side lengths, then use the cosine rule to find the smallest angle.

   1. Label the vectors. Let

   $\vec{a} = 2\vec{i} + \vec{j} - 2\vec{k}, \quad \vec{b} = 2\sqrt{3}\,\vec{i} - 2\sqrt{3}\,\vec{j} + \sqrt{3}\,\vec{k}.$

   These are two sides. The third side is

   $\vec{c} = \vec{b} - \vec{a}.$

   2. Find the magnitudes (side lengths).

   $|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3.$

   $|\vec{b}| = \sqrt{(2\sqrt{3})^2 + (-2\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{12 + 12 + 3} = \sqrt{27} = 3\sqrt{3}.$

   Now compute $\vec{c}$:

   $\vec{c} = (2\sqrt{3} - 2)\vec{i} + (-2\sqrt{3} - 1)\vec{j} + (\sqrt{3} + 2)\vec{k}.$

   Its magnitude:

   $|\vec{c}|^2 = (2\sqrt{3} - 2)^2 + (-2\sqrt{3} - 1)^2 + (\sqrt{3} + 2)^2.$

   Expand each:

   • $(2\sqrt{3} - 2)^2 = 4\cdot3 - 8\sqrt{3} + 4 = 12 - 8\sqrt{3} + 4 = 16 - 8\sqrt{3}$.
   • $(-2\sqrt{3} - 1)^2 = 4\cdot3 + 4\sqrt{3} + 1 = 12 + 4\sqrt{3} + 1 = 13 + 4\sqrt{3}$.
   • $(\sqrt{3} + 2)^2 = 3 + 4\sqrt{3} + 4 = 7 + 4\sqrt{3}$.

   Sum:

   $|\vec{c}|^2 = (16 - 8\sqrt{3}) + (13 + 4\sqrt{3}) + (7 + 4\sqrt{3}) = 36 + 0\sqrt{3} = 36.$

   So $|\vec{c}| = 6$.

   The three side lengths are $3$, $3\sqrt{3}$, and $6$.

   3. Identify the smallest side and the least angle.

      Compare the three lengths: $3\sqrt{3} \approx 5.196$, so the order is $3 < 3\sqrt{3} < 6$.

      The smallest side is $3$, opposite the smallest angle. That angle lies between the other two sides: $3\sqrt{3}$ and $6$.

      Use the cosine rule:

   $\cos(\text{smallest angle}) = \frac{(3\sqrt{3})^2 + 6^2 - 3^2}{2 \cdot 3\sqrt{3} \cdot 6}.$

   Compute:

   $(3\sqrt{3})^2 = 27,\quad 6^2 = 36,\quad 3^2 = 9.$

   Numerator: $27 + 36 - 9 = 54$.

   Denominator: $2 \cdot 3\sqrt{3} \cdot 6 = 36\sqrt{3}$.

   So

   $\cos\theta = \frac{54}{36\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}.$

   Hence $\theta = \frac{\pi}{6}$.

   4. Find the perimeter. Perimeter = $3 + 3\sqrt{3} + 6 = 9 + 3\sqrt{3}$.

   Watch out

   A common mistake is to assume the given vectors themselves are the sides of the triangle without checking which is smallest. Here $3\sqrt{3} \approx 5.196$ is actually smaller than $6$, so the ordering matters for the cosine rule.

   ✓Final answer

   The least angle is $\frac{\pi}{6}$ and the perimeter is $9 + 3\sqrt{3}$, which corresponds to option (A).

6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If $\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{b} = 9\hat{i} + 6\hat{j} - 18\hat{k}$ are two vectors, then $\dfrac{\text{Projection of } \vec{b} \text{ on } \vec{a}}{\text{Projection of } \vec{a} \text{ on } \vec{b}} =$ (A) $21$ (B) $7$ (C) $\dfrac{7}{3}$ (D) $3$

   ›Reveal solutionSolution

   The ratio of the projection of $\vec{b}$ on $\vec{a}$ to the projection of $\vec{a}$ on $\vec{b}$ simplifies to the ratio of the magnitudes of the two vectors. Computing the magnitudes gives $\frac{|\vec{b}|}{|\vec{a}|} = \frac{21}{3} = 7$, so the answer is (B).

   The key idea is that the projection of one vector onto another is not symmetric:

   $\text{proj}_{\vec{a}}(\vec{b}) = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|}, \quad \text{proj}_{\vec{b}}(\vec{a}) = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}.$

   So the ratio of the two projections is

   $\frac{\text{proj}_{\vec{a}}(\vec{b})}{\text{proj}_{\vec{b}}(\vec{a})} = \frac{\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|}}{\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}} = \frac{|\vec{b}|}{|\vec{a}|}.$

   The dot product cancels out entirely — we only need the lengths of the vectors.

   1. Find $|\vec{a}|$ $\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$

   $|\vec{a}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.$

   2. Find $|\vec{b}|$ $\vec{b} = 9\hat{i} + 6\hat{j} - 18\hat{k}$

   $|\vec{b}| = \sqrt{9^2 + 6^2 + (-18)^2} = \sqrt{81 + 36 + 324} = \sqrt{441} = 21.$

   3. Take the ratio

   $\frac{|\vec{b}|}{|\vec{a}|} = \frac{21}{3} = 7.$

   Tip

   Notice that the dot product $\vec{a}\cdot\vec{b} = 9 - 12 - 36 = -39$ is negative, meaning the vectors point in roughly opposite directions. But projection uses absolute length (scalar projection), so the sign doesn't affect the ratio — it cancels out anyway.

   Watch out

   A common mistake is to compute the vector projection (which includes a unit vector direction) instead of the scalar projection. The problem asks for "projection" in the scalar sense (the length of the shadow), so we use $\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|}$, not $\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\vec{a}$.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{b} = 9\hat{i} + 6\hat{j} - 18\hat{k}$ are two vectors, then $\dfrac{\text{Projection of } \vec{b} \text{ on } \vec{a}}{\text{Projection of } \vec{a} \text{ on } \vec{b}} =$ (A) $3$ (B) $7$ (C) $\dfrac{7}{3}$ (D) $21$

   ›Reveal solutionSolution

   The ratio of the projection of $\vec{b}$ on $\vec{a}$ to the projection of $\vec{a}$ on $\vec{b}$ equals the ratio of the magnitudes of the two vectors, which simplifies to $7$. The correct option is (B).

   ---

   Concept & Intuition

   The projection of one vector onto another measures how much of the first vector lies along the direction of the second.

   If you think of $\vec{a}$ and $\vec{b}$ as arrows, the projection of $\vec{b}$ onto $\vec{a}$ is the length of the shadow $\vec{b}$ casts on the line of $\vec{a}$.

   The formula for the scalar projection of $\vec{b}$ onto $\vec{a}$ is

   $\text{proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$

   and similarly, the projection of $\vec{a}$ onto $\vec{b}$ is

   $\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}.$

   Notice that both projections share the same dot product in the numerator. So when we take their ratio, the dot product cancels out, leaving only the ratio of the magnitudes. That’s the key insight — we don’t even need to compute the dot product explicitly.

   ---

   Step-by-step solution

   1. Write the projection formulas

   $\text{proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}, \qquad \text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}.$

   2. Form the required ratio

   $\frac{\text{proj}_{\vec{a}} \vec{b}}{\text{proj}_{\vec{b}} \vec{a}} = \frac{\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}}{\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}} = \frac{|\vec{b}|}{|\vec{a}|}.$

   The dot product cancels (provided it is nonzero — here it is, as we’ll see).

   3. Compute the magnitudes For $\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$:

   $|\vec{a}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.$

   For $\vec{b} = 9\hat{i} + 6\hat{j} - 18\hat{k}$:

   $|\vec{b}| = \sqrt{9^2 + 6^2 + (-18)^2} = \sqrt{81 + 36 + 324} = \sqrt{441} = 21.$

   4. Find the ratio

   $\frac{|\vec{b}|}{|\vec{a}|} = \frac{21}{3} = 7.$

   Tip

   You never needed to compute $\vec{a} \cdot \vec{b}$ — it cancels. This shortcut works whenever both projections are nonzero.

   Watch out

   A common mistake is to compute the vector projection instead of the scalar projection. The scalar projection is a length (no direction), and that’s what the problem asks for. The vector projection would have a unit vector attached, changing the ratio.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.Two adjacent sides of a triangle are represented by the vectors $2\vec{i} + \vec{j} - 2\vec{k}$ and $2\sqrt{3}\vec{i} - 2\sqrt{3}\vec{j} + \sqrt{3}\vec{k}$. Then the least angle of the triangle and perimeter of the triangle are respectively (A) $\dfrac{\pi}{3}; 3(3+\sqrt{3})$ (B) $\dfrac{\pi}{12}; 6+3\sqrt{2}$ (C) $\dfrac{\pi}{2}; 12$ (D) $\dfrac{\pi}{6}; 9+3\sqrt{3}$

   ›Reveal solutionSolution

   The triangle's two given sides have lengths 3 and $3\sqrt3$; the third side (their vector difference) has length 6. The least angle — opposite the shortest side — is $\dfrac{\pi}{6}$, and the perimeter is $9+3\sqrt3$. The correct option is (D).

   ---

   Two adjacent sides of a triangle are given by

   $\vec{a} = 2\vec{i} + \vec{j} - 2\vec{k}, \qquad \vec{b} = 2\sqrt{3}\,\vec{i} - 2\sqrt{3}\,\vec{j} + \sqrt{3}\,\vec{k}.$

   If both start from the same vertex $A$ (with $B$ and $C$ the tips of $\vec a$ and $\vec b$ respectively), then $AB=\vec a$, $AC=\vec b$, and the third side is $\vec{c}=\vec{b}-\vec{a}$ (from $B$ to $C$).

   Concept & Intuition:

   In any triangle, the smallest angle is opposite the shortest side. So we compute all three side lengths, identify the shortest, then find the angle opposite it (using the dot product / law of cosines). The perimeter is simply the sum of the three lengths.

   ---

   1. Lengths of the given sides

   $|\vec{a}| = \sqrt{2^2+1^2+(-2)^2} = \sqrt{9} = 3.$

   $|\vec{b}| = \sqrt{(2\sqrt3)^2+(-2\sqrt3)^2+(\sqrt3)^2} = \sqrt{12+12+3} = \sqrt{27} = 3\sqrt3.$

   2. Third side vector and its length

   $\vec{c}=\vec b-\vec a = (2\sqrt3-2)\vec i + (-2\sqrt3-1)\vec j + (\sqrt3+2)\vec k.$

   $|\vec c|^2 = (2\sqrt3-2)^2+(-2\sqrt3-1)^2+(\sqrt3+2)^2.$

   • $(2\sqrt3-2)^2 = 12-8\sqrt3+4 = 16-8\sqrt3$
   • $(-2\sqrt3-1)^2 = 12+4\sqrt3+1 = 13+4\sqrt3$
   • $(\sqrt3+2)^2 = 3+4\sqrt3+4 = 7+4\sqrt3$

   Sum: $(16-8\sqrt3)+(13+4\sqrt3)+(7+4\sqrt3) = 36$. So $|\vec c| = \sqrt{36}=6$.

   The three side lengths are $AB=3$, $AC=3\sqrt3\approx5.2$, $BC=6$.

   3. Identify the least angle

   The shortest side is $AB=3$, so the least angle is the one opposite it, at vertex $C$ — the angle between $CA=-\vec b$ and $CB=-\vec c$, which is the same as the angle between $\vec b$ and $\vec c$.

   4. Angle between $\vec b$ and $\vec c$

   $\vec b\cdot\vec c = 2\sqrt3(2\sqrt3-2) + (-2\sqrt3)(-2\sqrt3-1) + \sqrt3(\sqrt3+2).$

   • $2\sqrt3(2\sqrt3-2) = 12-4\sqrt3$
   • $(-2\sqrt3)(-2\sqrt3-1) = 12+2\sqrt3$
   • $\sqrt3(\sqrt3+2) = 3+2\sqrt3$

   Sum: $(12-4\sqrt3)+(12+2\sqrt3)+(3+2\sqrt3) = 27$.

   $\cos\theta = \frac{\vec b\cdot\vec c}{|\vec b||\vec c|} = \frac{27}{(3\sqrt3)(6)} = \frac{27}{18\sqrt3} = \frac{3}{2\sqrt3} = \frac{\sqrt3}{2}.$

   So $\theta = \dfrac{\pi}{6}$ — this is the least angle (confirmed by the law of cosines: $AB^2=AC^2+BC^2-2\cdot AC\cdot BC\cos C \Rightarrow 9=27+36-36\sqrt3\cos C \Rightarrow \cos C=\frac{54}{36\sqrt3}=\frac{\sqrt3}{2}$).

   5. Perimeter

   $\text{Perimeter} = 3+3\sqrt3+6 = 9+3\sqrt3.$

   Tip

   Always sanity-check the triangle inequality: $3+3\sqrt3\approx8.2>6$, $3+6>3\sqrt3$, and $3\sqrt3+6>3$ — all hold, so this is a valid triangle.

   Watch out

   A common mistake is to take the angle between $\vec a$ and $\vec b$ (at vertex $A$) as the least angle — but that angle is opposite the largest side $BC=6$, making it the largest angle, not the smallest.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.The angle between force $\vec{F} = 3\hat{i} + 4\hat{j} - 5\hat{k}$ and displacement $\vec{d} = 5\hat{i} + 4\hat{j} + 3\hat{k}$ is (A) $\cos^{-1}(0.16)$ (B) $\cos^{-1}(0.32)$ (C) $\cos^{-1}(0.24)$ (D) $\cos^{-1}(0.64)$

   ›Reveal solutionSolution

   The angle between two vectors comes from $\cos\theta=\dfrac{\vec F\cdot\vec d}{|\vec F||\vec d|}$. Here $\vec F\cdot\vec d=16$ and $|\vec F|=|\vec d|=5\sqrt2$, so $\cos\theta=0.32$.

   Solution

   $\vec F=3\hat i+4\hat j-5\hat k,\qquad \vec d=5\hat i+4\hat j+3\hat k.$

   Dot product:

   $\vec F\cdot\vec d=(3)(5)+(4)(4)+(-5)(3)=15+16-15=16.$

   Magnitudes:

   $|\vec F|=\sqrt{3^2+4^2+(-5)^2}=\sqrt{9+16+25}=\sqrt{50}=5\sqrt2,$

   $|\vec d|=\sqrt{5^2+4^2+3^2}=\sqrt{25+16+9}=\sqrt{50}=5\sqrt2.$

   Angle:

   $\cos\theta=\frac{\vec F\cdot\vec d}{|\vec F||\vec d|}=\frac{16}{5\sqrt2\cdot 5\sqrt2}=\frac{16}{50}=0.32.$

   Hence $\theta=\cos^{-1}(0.32)$.

   ✓Final answer

   $\theta=\cos^{-1}(0.32)$, option (B).

10. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.Let $A = (3, 4, 0)$, $B = (4, 4, 4)$, $C = (-6, 2, 3)$ and $D = (1, 1, 2)$. If $\theta$ is the acute angle between the lines $AB$ and $CD$ then $\cos \theta =$ (A) $\dfrac{4}{17\sqrt{3}}$ (B) $\dfrac{3}{17\sqrt{3}}$ (C) $\dfrac{12}{17\sqrt{3}}$ (D) $\dfrac{11}{17\sqrt{3}}$

    ›Reveal solutionSolution

    To find the cosine of the acute angle between two lines, we first determine their direction vectors. Then, we use the dot product formula, taking the absolute value of the dot product to ensure we get the acute angle. The result is $\boxed{\dfrac{3}{17\sqrt{3}}}$.

    The angle between two lines in 3D space is defined as the angle between their direction vectors. If the lines are $L_1$ with direction vector $\vec{d_1}$ and $L_2$ with direction vector $\vec{d_2}$, the cosine of the angle $\theta$ between them is given by the dot product formula.

    The dot product of two vectors $\vec{a}$ and $\vec{b}$ is related to the angle $\theta$ between them by the formula $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Rearranging this, we get $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.

    Since we are looking for the acute angle, we must ensure that $\cos \theta$ is positive. If the direct calculation of $\frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$ yields a negative value, it means the angle is obtuse. To get the acute angle, we simply take the absolute value of the dot product in the numerator.

    The cosine of the acute angle $\theta$ between two lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$ is given by:

    $\cos \theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}| |\vec{d_2}|}$

    Let's apply this concept to the given points.

    1. Determine the direction vector of line AB.

       The line AB passes through points $A = (3, 4, 0)$ and $B = (4, 4, 4)$.

       The direction vector $\vec{d_1}$ can be found by subtracting the coordinates of A from B:

       $\vec{d_1} = \vec{AB} = B - A = (4-3, 4-4, 4-0) = (1, 0, 4)$.

    2. Determine the direction vector of line CD.

       The line CD passes through points $C = (-6, 2, 3)$ and $D = (1, 1, 2)$.

       The direction vector $\vec{d_2}$ can be found by subtracting the coordinates of C from D:

       $\vec{d_2} = \vec{CD} = (1-(-6), 1-2, 2-3) = (7, -1, -1)$.

    3. Calculate the dot product of the direction vectors.

       The dot product $\vec{d_1} \cdot \vec{d_2}$ is calculated as the sum of the products of corresponding components:

       $\vec{d_1} \cdot \vec{d_2} = (1)(7) + (0)(-1) + (4)(-1)$

       $\vec{d_1} \cdot \vec{d_2} = 7 + 0 - 4 = 3$.

    4. Calculate the magnitudes of the direction vectors.

       The magnitude of a vector $(x, y, z)$ is $\sqrt{x^2 + y^2 + z^2}$.

       $|\vec{d_1}| = \sqrt{1^2 + 0^2 + 4^2} = \sqrt{1 + 0 + 16} = \sqrt{17}$.

       $|\vec{d_2}| = \sqrt{7^2 + (-1)^2 + (-1)^2} = \sqrt{49 + 1 + 1} = \sqrt{51}$.

    5. Substitute the values into the formula for $\cos \theta$.

       Since we need the acute angle, we use the absolute value of the dot product. In this case, the dot product is already positive ($3$), so the absolute value does not change it.

       $\cos \theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}| |\vec{d_2}|} = \frac{|3|}{\sqrt{17} \cdot \sqrt{51}}$.

       $\cos \theta = \frac{3}{\sqrt{17 \times 51}}$.

       We can simplify the denominator: $\sqrt{51} = \sqrt{3 \times 17}$.

       So, $\cos \theta = \frac{3}{\sqrt{17 \times 3 \times 17}} = \frac{3}{\sqrt{17^2 \times 3}} = \frac{3}{17\sqrt{3}}$.

    ✓Final answer

    The value of $\cos \theta$ is $\boxed{\dfrac{3}{17\sqrt{3}}}$.

11. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
    Q.Find the component of vector $\vec{P} = 2\hat{i} + 3\hat{j}$ along the direction of vector $\vec{Q} = \hat{i} + \hat{j}$. (A) $2$ (B) $2\sqrt{5}$ (C) $\dfrac{5}{\sqrt{2}}$ (D) $\dfrac{\sqrt{2}}{5}$

    ›Reveal solutionSolution

    The component of a vector along another direction is found by projecting it onto that direction — the scalar projection formula $\frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|}$ gives the answer $\frac{5}{\sqrt{2}}$, which is option (C).

    The idea is simple: when we ask for the "component" of $\vec{P}$ along $\vec{Q}$, we mean the length of the shadow $\vec{P}$ casts onto the line of $\vec{Q}$. That's the scalar projection — not a vector, just a signed magnitude. The formula comes directly from the dot product: $\vec{P} \cdot \vec{Q} = |\vec{P}| |\vec{Q}| \cos \theta$, and $|\vec{P}| \cos \theta$ is exactly that component. So we compute $\frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|}$.

    Let's work it through.

    1. Write the vectors clearly

       $\vec{P} = 2\hat{i} + 3\hat{j}$

       $\vec{Q} = \hat{i} + \hat{j}$

    2. Compute the dot product

       $\vec{P} \cdot \vec{Q} = (2)(1) + (3)(1) = 2 + 3 = 5$

    3. Find the magnitude of $\vec{Q}$

       $|\vec{Q}| = \sqrt{1^2 + 1^2} = \sqrt{2}$

    4. Apply the projection formula

       Component of $\vec{P}$ along $\vec{Q}$ = $\frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|} = \frac{5}{\sqrt{2}}$

    Watch out

    A common mistake is to divide by $|\vec{P}|$ instead of $|\vec{Q}|$, or to forget the magnitude in the denominator entirely. The dot product alone gives a scalar, but it's not the component — you must divide by the length of the direction vector.

    Tip

    If $\vec{Q}$ were a unit vector, the component would just be the dot product. Here $\vec{Q}$ isn't unit, so we normalize it first: $\hat{Q} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$, then $\vec{P} \cdot \hat{Q} = \frac{5}{\sqrt{2}}$ — same result, fewer steps.

    ✓Final answer

    The component is $\boxed{\frac{5}{\sqrt{2}}}$, which corresponds to option (C).

12. TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
    Q.The square of resultant of two equal forces is three times their product. Angle between the forces is? (A) $\pi$ (B) $\pi/2$ (C) $\pi/3$ (D) $\pi/4$

    ›Reveal solutionSolution

    The problem reduces to using the resultant formula for two vectors: $R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta$. Given equal forces and $R^2 = 3F^2$, solving gives $\cos\theta = 1/2$, so $\theta = \pi/3$. The correct option is (C).

    The key concept here is the parallelogram law of vector addition. When two forces (vectors) act at a point, the magnitude of their resultant depends on the angle between them. The problem gives a relationship between the square of the resultant and the product of the forces, which directly leads to an equation for the cosine of the angle.

    Why this approach works:

    We have two equal forces, say each of magnitude $F$. Their resultant $R$ satisfies $R^2 = F^2 + F^2 + 2F \cdot F \cos\theta = 2F^2(1 + \cos\theta)$. The problem states $R^2 = 3 \times (\text{product of the forces}) = 3F^2$. Equating these gives a simple trigonometric equation.

    Step-by-step solution:

    1. Write the resultant formula for two equal forces. For two forces of equal magnitude $F$ with an angle $\theta$ between them, the magnitude of the resultant $R$ is given by:

    $R^2 = F^2 + F^2 + 2F \cdot F \cos\theta = 2F^2(1 + \cos\theta).$

    2. Translate the given condition into an equation. The problem says: "The square of resultant of two equal forces is three times their product." Their product is $F \times F = F^2$, so:

    $R^2 = 3F^2.$

    3. Set the two expressions for $R^2$ equal.

    $2F^2(1 + \cos\theta) = 3F^2.$

    Since $F \neq 0$, divide both sides by $F^2$:

    $2(1 + \cos\theta) = 3.$

    4. Solve for $\cos\theta$.

    $1 + \cos\theta = \frac{3}{2} \quad \Rightarrow \quad \cos\theta = \frac{1}{2}.$

    5. Find the angle. The angle whose cosine is $1/2$ in the range $[0, \pi]$ (the usual range for the angle between vectors) is:

    $\theta = \frac{\pi}{3}.$

    Tip

    A common mistake is to forget the factor of 2 in the resultant formula or to misinterpret "product" as the dot product. Here "product" means the scalar product of the magnitudes, i.e., $F \times F$.

    Watch out

    If you mistakenly use $R^2 = F^2 + F^2$ (the Pythagorean case), you'd get $2F^2 = 3F^2$, which is impossible. That would lead you to think the angle is $\pi/2$, but that's a trap — the correct formula includes the $2F^2\cos\theta$ term.

    ✓Final answer

    The correct option is (C).

    ANSWER: C