Q.The angle between $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$ is

Showing the 12 most recent of 16 on this concept.

1. 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

2. 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

3. 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

4. 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).

5. 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).

6. 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).

7. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.It is found that a non-zero current element is unable to produce any magnetic field at a particular point. Then the angle between the current element and the position vector of that point with respect to the current element is (A) must be $45^\circ$ (B) may be $0^\circ$ or $180^\circ$ (C) must be $90^\circ$ (D) may be $30^\circ$ or $60^\circ$

   ›Reveal solutionSolution

   The magnetic field from a current element is zero when the cross product of the current element vector and the position vector vanishes — this happens when the angle between them is $0^\circ$ or $180^\circ$, so the correct option is (B).

   The key concept here is the Biot–Savart law for a current element. It tells us that the magnetic field produced by a small segment of current-carrying wire depends on the cross product of the current element vector $d\vec{l}$ and the position vector $\vec{r}$ from the element to the point of interest. If that cross product is zero, the magnetic field is zero — regardless of the size of the current element.

   Why does this happen? Because the cross product $d\vec{l} \times \vec{r}$ has magnitude $|d\vec{l}|\,|\vec{r}|\,\sin\theta$, where $\theta$ is the angle between the directions of $d\vec{l}$ and $\vec{r}$. When $\sin\theta = 0$, the cross product vanishes, and so does the magnetic field. So the question reduces to: for what angles is $\sin\theta = 0$?

   Let’s work through it step by step.

   1. Recall the Biot–Savart law for a current element The magnetic field $d\vec{B}$ at a point due to a current element $I\,d\vec{l}$ is

   $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I\,d\vec{l} \times \hat{r}}{r^2}$

   where $\hat{r}$ is the unit vector from the element to the point. The magnitude is

   $|d\vec{B}| = \frac{\mu_0}{4\pi} \frac{I\,|d\vec{l}| \sin\theta}{r^2}$

   with $\theta$ the angle between $d\vec{l}$ and $\vec{r}$.

   2. Condition for zero magnetic field

      For $|d\vec{B}|$ to be zero, we need $\sin\theta = 0$ (since $I$, $|d\vec{l}|$, and $r$ are non-zero).

      $\sin\theta = 0$ when $\theta = 0^\circ$ or $\theta = 180^\circ$ (or any integer multiple of $180^\circ$, but within $0^\circ$ to $180^\circ$ these are the only possibilities).

   3. Interpretation of the angles

      • At $\theta = 0^\circ$, the current element points directly toward the point.
      • At $\theta = 180^\circ$, the current element points directly away from the point. In both cases, the direction of $d\vec{l}$ is exactly along the line joining the element to the point, so the cross product is zero.

   4. Check the options

      • (A) $45^\circ$ gives $\sin45^\circ \neq 0$, so field is non-zero.
      • (B) $0^\circ$ or $180^\circ$ — exactly the condition we found.
      • (C) $90^\circ$ gives $\sin90^\circ = 1$, maximum field, not zero.
      • (D) $30^\circ$ or $60^\circ$ both give non-zero sine values.

   Watch out

   A common mistake is to think the field is zero when the current element is perpendicular to the position vector — that actually gives the maximum field, not zero. Zero happens only when they are parallel or anti-parallel.

   Tip

   You can remember this as: "No field when the current points straight at or away from the point." The cross product vanishes when the two vectors are collinear.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. 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

9. 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).

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 2022Set eng-2022-07-19-FN1 markMCQ
    Q.An aircraft is flying at a height of $'H'$ above the ground and at a speed of $'V'$. The maximum angle subtended at a ground observation point by the aircraft after time $T$ is (A) $\tan^{-1} \left( \dfrac{VT}{H} \right)$ (B) $\tan^{-1} \left( \dfrac{VT}{2H} \right)$ (C) $2 \tan^{-1} \left( \dfrac{2VT}{H} \right)$ (D) $2 \tan^{-1} \left( \dfrac{VT}{2H} \right)$

    ›Reveal solutionSolution

    The maximum angle subtended by a horizontal line segment (the aircraft's path) at a point on a parallel line (the ground) occurs when the observation point is directly below the midpoint of the segment. Using trigonometry, this maximum angle is $\boxed{2 \tan^{-1} \left( \dfrac{VT}{2H} \right)}$.

    The problem asks for the maximum angle subtended by the aircraft's path at any point on the ground. This means we need to consider the aircraft's movement over time $T$ as a line segment and then find the optimal position for a ground observer to see this segment under the largest possible angle.

    Concept and Intuition

    1. Aircraft's Path as a Segment: The aircraft flies at a constant speed $V$ for a time $T$. During this time, it covers a horizontal distance $L = VT$. Since it flies at a constant height $H$, its path is a horizontal line segment of length $L$ at height $H$ above the ground.
    2. Maximizing Subtended Angle: Imagine a fixed line segment $AB$ (the aircraft's path) and a line $l$ parallel to $AB$ (the ground). We want to find a point $P$ on $l$ such that the angle $\angle APB$ is maximized. Geometrically, this maximum angle occurs when the observation point $P$ is directly below the midpoint of the segment $AB$. Any other position for $P$ would result in a smaller angle. This is because, for a fixed segment, the locus of points subtending a constant angle is a circular arc. To maximize the angle, we need to find the circle that passes through $A$ and $B$ and is tangent to the line $l$. The point of tangency will be directly below the midpoint of $AB$.

    Let's apply this understanding step-by-step.

    1. Determine the length of the aircraft's path segment:

       The aircraft flies at a speed $V$ for a time $T$. The horizontal distance it covers is:

       $L = V \times T$

       This segment of length $L$ is at a constant height $H$ above the ground.

    2. Identify the optimal observation point:

       To maximize the angle subtended by this horizontal segment at a point on the ground, the observation point must be directly below the midpoint of the segment.

       Let's set up a coordinate system for clarity. We can place the observation point $P$ at the origin $(0, 0)$ on the ground.

       Since $P$ is directly below the midpoint of the aircraft's path, the midpoint of the path will be at $(0, H)$.

       The aircraft's path segment, of total length $L$, will therefore extend from $x = -L/2$ to $x = L/2$ at height $H$.

       So, the initial position of the aircraft is $A = (-L/2, H)$.

       The final position of the aircraft is $B = (L/2, H)$.

    3. Calculate the maximum angle subtended:

       We need to find the angle $\theta = \angle APB$.

       Consider the triangle $\triangle APB$. Since the observation point $P$ is directly below the midpoint of $AB$, the triangle $\triangle APB$ is isosceles with $PA = PB$.

       Let $M$ be the midpoint of $AB$. So $M = (0, H)$.

       The line segment $PM$ is perpendicular to $AB$. This means $\triangle APM$ is a right-angled triangle.

       The angle $\angle APB$ is bisected by $PM$, so $\angle APB = 2 \angle APM$.

       In the right-angled triangle $\triangle APM$:

       • The side $PM$ is the height $H$.
       • The side $AM$ is half the length of the segment $AB$, so $AM = L/2$.
       • Substituting $L = VT$, we get $AM = VT/2$.

       Now, we can use the tangent function to find $\angle APM$:

    $\tan(\angle APM) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AM}{PM}$

    $\tan(\angle APM) = \frac{VT/2}{H} = \frac{VT}{2H}$

    Therefore, the angle $\angle APM$ is:
    

    $\angle APM = \tan^{-1} \left( \frac{VT}{2H} \right)$

    The maximum angle subtended by the aircraft's path at the ground observation point is $\theta = 2 \angle APM$:
    

    $\theta = 2 \tan^{-1} \left( \frac{VT}{2H} \right)$

    ✓Final answer

    The maximum angle subtended at a ground observation point by the aircraft after time $T$ is $\boxed{2 \tan^{-1} \left( \dfrac{VT}{2H} \right)}$.

12. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
    Q.An ant starts from the origin and crawls 10 cm along the x-axis and then 20 cm along the y-axis. The dot product of the ant's displacement vector with the position vector of a point that makes $45^\circ$ with the x-axis and has a magnitude of $\sqrt{2}$ cm is (A) 30 cm (B) $30\sqrt{2}$ cm (C) $\dfrac{30}{\sqrt{2}}$ cm (D) 15 cm

    ›Reveal solutionSolution

    The dot product is computed by summing the products of corresponding components of the two vectors. The ant’s displacement is (10, 20) and the given position vector is (1, 1), so the dot product is 10·1 + 20·1 = 30. The correct option is (A).

    The key idea is that the dot product of two vectors is simply the sum of the products of their components. No trigonometry is needed here because the position vector’s magnitude and angle are given only to help you find its components — but once you have those, the calculation is straightforward.

    Why this approach works:

    The dot product measures how much one vector “projects” onto another. If you know the components of both vectors, you can compute it directly. The problem gives the ant’s displacement in components (10 along x, 20 along y) and describes the second vector by its magnitude and direction — so you first find its components, then multiply and add.

    Step-by-step solution:

    1. Ant’s displacement vector The ant moves 10 cm along the x-axis, then 20 cm along the y-axis. So its displacement vector is

    $\vec{A} = (10, 20) \text{ cm}.$

    2. The second vector’s components The position vector makes a $45^\circ$ angle with the x-axis and has magnitude $\sqrt{2}$ cm. Its components are:

    $\vec{B} = (\sqrt{2} \cos 45^\circ,\ \sqrt{2} \sin 45^\circ).$

    Since $\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}$, we get:

    $\vec{B} = \left( \sqrt{2} \cdot \frac{1}{\sqrt{2}},\ \sqrt{2} \cdot \frac{1}{\sqrt{2}} \right) = (1, 1) \text{ cm}.$

    3. Compute the dot product The dot product is:

    $\vec{A} \cdot \vec{B} = (10)(1) + (20)(1) = 10 + 20 = 30.$

    The units are cm² (since both vectors are in cm), but the answer choices simply say “cm” — this is a common convention in such problems.

    Watch out

    A common mistake is to try to use the formula $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta$ without first finding the angle between the two vectors. Here, the angle between $\vec{A}$ and $\vec{B}$ is not $45^\circ$ — that’s the angle $\vec{B}$ makes with the x-axis, not the angle between the vectors. Using that formula would require extra work; component method is simpler.

    Tip

    Whenever a vector is given by magnitude and direction, immediately convert to components. It almost always makes dot products trivial.

    ✓Final answer

    The correct option is (A).

    ANSWER: A