For two vectors $\vec{A}$ and $\vec{B}$, $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ is always true when (Note: more than one of the given options may be correct.) (a) $|\vec{A}| = |\vec{B}| \neq 0$ (b) $\vec{A} \perp \vec{B}$ (c) $|\vec{A}| = |\vec{B}| \neq 0$ and $\vec{A}$ and $\vec{B}$ are parallel or anti parallel (d) when either $|\vec{A}|$ or $|\vec{B}|$ is zero.

> [!TLDR] > The condition $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ boils down to $\vec{A} \cdot \vec{B} = 0$, i.e., the vectors are perpendicular. This holds when $\vec{A} \perp \vec{B}$ **or** when either vector is zero (since a zero vector is trivially perpendicular to any vector). So the correct options are **(B)** and **(D)**. The key is to avoid memorizing — instead, square both magnitudes and see what the equality forces. **Why the Triangle Inequality idea?** The magnitudes $|\vec{A} + \vec{B}|$ and $|\vec{A} - \vec{B}|$ are the lengths of the diagonals of the parallelogram formed by $\vec{A}$ and $\vec{B}$. For these diagonals to be equal, the parallelogram must be a rectangle — meaning the sides are perpendicular. That’s the geometric intuition. Algebraically, squaring removes the square root and gives a clean dot-product condition. 1. **Square both sides** Since magnitudes are non-negative, $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ is equivalent to $$|\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2.$$ 2. **Expand using the dot product** Recall $|\vec{V}|^2 = \vec{V} \cdot \vec{V}$. So: $$(\vec{A} + \vec{B})\cdot(\vec{A} + \vec{B}) = (\vec{A} - \vec{B})\cdot(\vec{A} - \vec{B}).$$ Expanding: $$\vec{A}\cdot\vec{A} + 2\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{B} = \vec{A}\cdot\vec{A} - 2\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{B}.$$ 3. **Cancel common terms** $|\vec{A}|^2$ and $|\vec{B}|^2$ appear on both sides, so they cancel, leaving: $$2\vec{A}\cdot\vec{B} = -2\vec{A}\cdot\vec{B}.$$ This simplifies to $4\vec{A}\cdot\vec{B} = 0$, i.e., $$\vec{A}\cdot\vec{B} = 0.$$ > [!FORMULA] > $$|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \iff \vec{A}\cdot\vec{B} = 0$$ 4. **Interpret the dot product condition** $\vec{A}\cdot\vec{B} = 0$ means the vectors are **perpendicular** (orthogonal). But there’s a special case: if either $\vec{A}$ or $\vec{B}$ is the zero vector, then $\vec{A}\cdot\vec{B} = 0$ holds trivially

For two vectors vecA and vecB, |vecA + vecB| = |ve – Class 11 Physics

NCERT Exemplar · Q15

\vec{A}

(a)

|\vec{A}| = |\vec{B}| \neq 0

(b)

\vec{A} \perp \vec{B}

(c)

|\vec{A}| = |\vec{B}| \neq 0

and

\vec{A}

and

\vec{B}

are parallel or anti parallel

(d) when either

|\vec{A}|

|\vec{B}|

is zero.

Telangana TsbieMCQ· 1mImportance★★★★★

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Concept understanding — Dot Product Angle

Finding the Angle Between Vectors

Suppose you have two arrows drawn from the same point. One question is unavoidable in geometry, physics, and mechanics: what is the angle between them? You could measure it with a protractor on paper, but that fails the moment the vectors live in 3D. The dot product gives you the angle by pure calculation.

The Core Idea

The scalar (dot) product of two vectors has two faces that describe the same number:

\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \qquad \text{(components)}

\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta \qquad \text{(geometry)}

The first is easy to compute from coordinates; the second hides the angle $\theta$ (with $0 \le \theta \le \pi$ ) between the vectors. Setting them equal and solving for $\cos\theta$ gives the master formula.

\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}, \qquad \theta = \cos^{-1}\!\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\right)

Why It Works

Both vectors have a fixed length, so the only thing the dot product can "vary" with is how aligned they are. When they point the same way, $\cos\theta = 1$ and the dot product is as large as possible, $|\vec{a}||\vec{b}|$ . When they are perpendicular, $\cos\theta = 0$ and the dot product vanishes. When they point opposite ways, $\cos\theta = -1$ . Dividing $\vec{a}\cdot\vec{b}$ by the two lengths simply strips away the size information and leaves behind a pure measure of alignment — exactly $\cos\theta$ .

Note

The sign of the dot product tells you the type of angle at a glance: positive $\Rightarrow$ acute, zero $\Rightarrow$ right angle, negative $\Rightarrow$ obtuse.

Using the Formula

For $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 0\hat{j} + 0\hat{k}$ :

\vec{a}\cdot\vec{b} = 1, \quad |\vec{a}| = 3, \quad |\vec{b}| = 1

$\cos\theta = \frac{1}{3\cdot 1} = \frac{1}{3} \;\Rightarrow\; \theta = \cos^{-1}\tfrac{1}{3} \approx 70.5^\circ$ …

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The condition $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ boils down to $\vec{A} \cdot \vec{B} = 0$ , i.e., the vectors are perpendicular. This holds when $\vec{A} \perp \vec{B}$ or when either vector is zero (since a zero vector is trivially perpendicular to any vector). So the correct options are (B) and (D).

The key is to avoid memorizing — instead, square both magnitudes and see what the equality forces.

Why the Triangle Inequality idea?

The magnitudes $|\vec{A} + \vec{B}|$ and $|\vec{A} - \vec{B}|$ are the lengths of the diagonals of the parallelogram formed by $\vec{A}$ and $\vec{B}$ . For these diagonals to be equal, the parallelogram must be a rectangle — meaning the sides are perpendicular. That’s the geometric intuition. Algebraically, squaring removes the square root and gives a clean dot-product condition.

Square both sides Since magnitudes are non-negative, $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ is equivalent to

|\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2.

Expand using the dot product Recall $|\vec{V}|^2 = \vec{V} \cdot \vec{V}$ . So:

(\vec{A} + \vec{B})\cdot(\vec{A} + \vec{B}) = (\vec{A} - \vec{B})\cdot(\vec{A} - \vec{B}).

Expanding:

\vec{A}\cdot\vec{A} + 2\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{B} = \vec{A}\cdot\vec{A} - 2\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{B}.

Cancel common terms $|\vec{A}|^2$ and $|\vec{B}|^2$ appear on both sides, so they cancel, leaving:

2\vec{A}\cdot\vec{B} = -2\vec{A}\cdot\vec{B}.

This simplifies to $4\vec{A}\cdot\vec{B} = 0$ , i.e.,

\vec{A}\cdot\vec{B} = 0.

|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \iff \vec{A}\cdot\vec{B} = 0

Interpret the dot product condition

$\vec{A}\cdot\vec{B} = 0$ means the vectors are perpendicular (orthogonal). But there’s a special case: if either $\vec{A}$ or $\vec{B}$ is the zero vector, then $\vec{A}\cdot\vec{B} = 0$ holds trivially (since $0 \cdot \vec{B} = 0$ ). A zero vector has no direction, so it’s considered perpendicular to every vector by convention.
Check each option …

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Concept: Equal Diagonals Mean a Rectangle -- a Classical Synthetic-Geometry Fact, No Coordinates or Dot Products

Method: The Parallelogram Law of Geometry (Diagonals Equal $\iff$ Rectangle)

Rather than expanding $|\vec A+\vec B|^2$ and $|\vec A-\vec B|^2$ via the dot product, this method uses a classical fact from Euclidean geometry about parallelograms directly: the two diagonals of a parallelogram are equal in length if and only if the parallelogram is a rectangle (equivalently, iff its adjacent sides are perpendicular). No components or dot products appear anywhere.

Setting up the parallelogram

Place $\vec A$ and $\vec B$ tail-to-tail at a common point $O$ , and complete the parallelogram $OPQR$ they generate (with $\vec A = \overrightarrow{OP}$ , $\vec B=\overrightarrow{OR}$ , and $Q$ the fourth vertex). The two diagonals of this parallelogram are the well-known vector-addition results:

\overrightarrow{OQ} = \vec A + \vec B \qquad(\text{the "long" diagonal, sum of adjacent sides})

\overrightarrow{RP} = \vec A - \vec B \qquad(\text{the "short" diagonal, difference of adjacent sides})

Applying the classical theorem

The given condition $|\vec A+\vec B|=|\vec A-\vec B|$ is exactly the statement that these two diagonals have equal length. By the theorem above, this happens if and only if $OPQR$ is a rectangle, i.e. if and only if the adjacent sides $\vec A$ and $\vec B$ meet at a right angle:

|\vec A+\vec B| = |\vec A-\vec B| \iff \vec A \perp \vec B

Checking the special (degenerate) case

A "rectangle" with one side of zero length is a degenerate limiting case (it collapses to a line segment) -- but the geometric theorem still holds trivially there: if $\vec B=\vec 0$ (or $\vec A=\vec 0$ ), the parallelogram itself degenerates, and both diagonals collapse to the same segment $\vec A$ (or $\vec B$ ), which are trivially equal in length. So the perpendicularity theorem, extended to this boundary case, also covers "either vector is zero."

Checking why (A) and (C) fail, using the same picture …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$
›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.

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

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

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

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

Take the ratio $\frac{|\vec{b}|}{|\vec{a}|} = \frac{21}{3} = 7.$ …
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TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$
›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

Write the projection formulas

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

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).

Compute the magnitudes For $\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$ :
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TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
$2\vec{i} + \vec{j} - 2\vec{k}$
›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 $\vec c$
$\vec b\cdot\vec c = 2\sqrt3(2\sqrt3-2) + (-2\sqrt3)(-2\sqrt3-1) + \sqrt3(\sqrt3+2).$

$2\sqrt3(2\sqrt3-2) = 12-4\sqrt3$

$(-2\sqrt3)(-2\sqrt3-1) = 12+2\sqrt3$

$\sqrt3(\sqrt3+2) = 3+2\sqrt3$ …
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$2\vec{i} + \vec{j} - 2\vec{k}$
›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.

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

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 $6$ .

Identify the smallest side and the least angle. Compare the three lengths: $3\sqrt{3} \approx 5.196$ , so the order is $3 < 3\sqrt{3} < 6$ . …
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TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$\overrightarrow{BC} = 2\hat{i} + \hat{j} + \hat{k}$
›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. …
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TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$\vec{F} = 3\hat{i} + 4\hat{j} - 5\hat{k}$
›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:

$|\vec F|=\sqrt{3^2+4^2+(-5)^2}=\sqrt{9+16+25}=\sqrt{50}=5\sqrt2,$ …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$45^\circ$
›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.

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 $\vec{r}$ .

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).

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. …
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TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
$\vec{A} = 4\hat{i} + 7\hat{j}$
›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.

Identify the components

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

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

Compute the magnitude

The magnitude is

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

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

Check the options

(A) uses $\sqrt{11}$ — that’s $4+7$ , not the magnitude. …
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TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$\pi_1$
›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

$\cos\theta=\frac{-6}{\sqrt{14}\,\sqrt6}=\frac{-6}{2\sqrt{21}}=-\frac{3}{\sqrt{21}}=-\frac{\sqrt{21}}{7}.$ …
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TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$A = (3, 4, 0)$
›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.
$\theta$
Let's apply this concept to the given points.

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:

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

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:

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

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:

$\vec{d_1} \cdot \vec{d_2} = (1)(7) + (0)(-1) + (4)(-1)$ …
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TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
$'H'$
›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

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.

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.

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.

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 $B = (L/2, H)$ .

Calculate the maximum angle subtended:

We need to find the angle $\theta = \angle APB$ . …
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TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$45^\circ$
›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:

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

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

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

Compute the dot product The dot product is: $\vec{A} \cdot \vec{B} = (10)(1) + (20)(1) = 10 + 20 = 30.$ …
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Q.For two vectors A⃗\vec{A}A and B⃗\vec{B}B, ∣A⃗+B⃗∣=∣A⃗−B⃗∣|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|∣A+B∣=∣A−B∣ is always true when (Note: more than one of the given options may be correct.)

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Q.For two vectors $\vec{A}$ and $\vec{B}$ , $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ is always true when (Note: more than one of the given options may be correct.)

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