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.

Andhra Pradesh BieapMCQ· 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 46 on this concept.

AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$\vec{a} = 4\vec{i} - \vec{j} + \alpha\vec{k}$
›Reveal solutionSolution
Setting up $\cos\theta=\vec a\cdot\vec b/(|\vec a||\vec b|)=-2/7$ gives a quadratic in $\alpha$ with roots $2$ and $15.5$ ; then $\alpha_1+2\alpha_2=33$ .

Concept and Intuition

Both vectors have the same magnitude expression in $\alpha$ (a nice simplification to notice first), which keeps the resulting equation a clean single-variable quadratic instead of something messier.

Step-by-Step Solution

$\vec a\cdot\vec b=4(1)+(-1)(\alpha)+\alpha(-4)=4-\alpha-4\alpha=4-5\alpha$ .

$|\vec a|=\sqrt{16+1+\alpha^2}=\sqrt{17+\alpha^2}$ and $|\vec b|=\sqrt{1+\alpha^2+16}=\sqrt{17+\alpha^2}$ — identical.

So $\cos\theta=\dfrac{4-5\alpha}{17+\alpha^2}=-\dfrac27$ .

Cross-multiply: $7(4-5\alpha)=-2(17+\alpha^2)\Rightarrow28-35\alpha=-34-2\alpha^2\Rightarrow2\alpha^2-35\alpha+62=0$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$60°$
›Reveal solutionSolution
Setting the resultant's magnitude equal to each vector's own magnitude in the law of cosines forces $\cos\theta=-1/2$ , i.e. $\theta=120°$ — option (C).

Concept and Intuition

For two vectors of equal magnitude $A=B=R$ (and the resultant also equal to $R$ ), the vector triangle formed is actually an equilateral triangle (all three sides equal), whose interior angle is $60°$ — but the angle between the vectors (the angle you'd measure if you place them tail-to-tail) is the exterior supplement, $180°-60°=120°$ . The law of cosines derivation below confirms this directly.

Step-by-Step Solution

Law of cosines for resultant: $R^2=A^2+B^2+2AB\cos\theta$ .

Given $A=B=R$ (resultant magnitude equals magnitude of each vector): $R^2=R^2+R^2+2R^2\cos\theta$ .

Divide throughout by $R^2$ : $1=1+1+2\cos\theta \Rightarrow 1=2+2\cos\theta$ . …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$ABC$
›Reveal solutionSolution
Find $|AB|$ from the position vectors, then apply the law of cosines at the known angle $A$ .

Concept and Intuition

Once we know two sides meeting at a vertex ( $AB$ and $AC$ ) and the included angle there ( $A$ ), the third side $BC$ is fixed by the law of cosines — position vectors are just a way of encoding the side length $AB$ .

Step-by-Step Solution

$A=(1,3,4)$ , $B=(2,1,2)$ , so $\overline{AB}=B-A=(1,-2,-2)$ , giving $|AB|=\sqrt{1^2+(-2)^2+(-2)^2}=\sqrt9=3$ .

We are given $|AC|=5$ and $\angle A = \pi/3$ (the angle between $AB$ and $AC$ at vertex $A$ ).

By the law of cosines in $\triangle ABC$ : $BC^2 = AB^2+AC^2-2\cdot AB\cdot AC\cos A$ .

$=3^2+5^2-2(3)(5)\cos(\pi/3) = 9+25-30\left(\dfrac12\right)=34-15=19$ . …
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AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$A(1, 0, 1)$
›Reveal solutionSolution
Use position vectors to get the two enclosed sides at each vertex, apply the vector dot-product formula for the cosine of the included angle, then combine the squares.

Concept and Intuition

The angle of a triangle at a vertex is the angle between the two vectors formed by that vertex to the other two vertices. Vector algebra gives this directly via $\cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$ , with no need to first find all three side lengths and invoke the cosine rule separately for each angle.

Step-by-Step Solution

Compute the vectors from $A$ : $\vec{AB}=B-A=(-1,1,-2)$ , $\vec{AC}=C-A=(-2,1,-1)$ .

$|\vec{AB}|=\sqrt{1+1+4}=\sqrt6$ , $|\vec{AC}|=\sqrt{4+1+1}=\sqrt6$ , and $\vec{AB}\cdot\vec{AC}=(-1)(-2)+(1)(1)+(-2)(-1)=2+1+2=5$ .

So $\cos A=\dfrac{5}{6}$ , giving $\cos^2A=\dfrac{25}{36}$ .

Compute the vectors from $B$ : $\vec{BA}=A-B=(1,-1,2)$ , $\vec{BC}=C-B=(-1,0,1)$ .

$|\vec{BA}|=\sqrt6$ , $|\vec{BC}|=\sqrt{1+0+1}=\sqrt2$ , and $\vec{BA}\cdot\vec{BC}=(1)(-1)+(-1)(0)+(2)(1)=1$ . …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\bar{a},\bar{b},\bar{c}$
›Reveal solutionSolution
Expanding $|\bar a+\bar b+\bar c|^2$ and using the two known angles isolates $\bar c.\bar a$ , giving the third angle as $\frac{\pi}{3}$ .

Concept and Intuition

The squared magnitude of a vector sum expands into the sum of squared magnitudes plus twice the pairwise dot products. With two of the three pairwise angles already known, this single scalar equation is enough to solve for the third dot product — and hence the third angle.

Step-by-Step Solution

Expand: $|\bar a+\bar b+\bar c|^2=|\bar a|^2+|\bar b|^2+|\bar c|^2+2(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a)$ .

Substitute known magnitudes: $69 = 4+9+25+2(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a) = 38+2(\ldots)$ .

So $\bar a.\bar b+\bar b.\bar c+\bar c.\bar a=\dfrac{69-38}{2}=\dfrac{31}{2}=15.5$ .

Compute $\bar a.\bar b=|\bar a||\bar b|\cos\dfrac{\pi}{3}=2\cdot3\cdot\dfrac12=3$ .

Compute $\bar b.\bar c=|\bar b||\bar c|\cos\dfrac{\pi}{3}=3\cdot5\cdot\dfrac12=7.5$ .

So $\bar c.\bar a=15.5-3-7.5=5$ . …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\bar{i}+\bar{j}-\bar{k}$
›Reveal solutionSolution
The angle between the two triangular faces sharing edge AB equals the angle between their normal vectors, computed via a pair of cross products as $\cos^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)$ .

Concept and Intuition

The dihedral angle between two planes meeting along a common edge can be found from the angle between their normal vectors (normals are perpendicular to their respective planes, so the angle between normals directly reflects the angle between the planes, up to sign conventions).

Step-by-Step Solution

Position vectors: $A=(1,1,-1)$ , $B=(1,-1,2)$ , $C=(1,-2,1)$ , $D=(2,1,1)$ .

Compute edge vectors from $A$ : $\overrightarrow{AB}=B-A=(0,-2,3)$ , $\overrightarrow{AC}=C-A=(0,-3,2)$ , $\overrightarrow{AD}=D-A=(1,0,2)$ .

Normal to face $ABC$ : $\bar n_1=\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\bar i&\bar j&\bar k\\0&-2&3\\0&-3&2\end{vmatrix} = \bar i[(-2)(2)-(3)(-3)] - \bar j[(0)(2)-(3)(0)]+\bar k[(0)(-3)-(-2)(0)] = \bar i(-4+9) - \bar j(0) + \bar k(0) = (5,0,0)$ .

Normal to face $ABD$ : $\bar n_2=\overrightarrow{AB}\times\overrightarrow{AD} = \begin{vmatrix}\bar i&\bar j&\bar k\\0&-2&3\\1&0&2\end{vmatrix} = \bar i[(-2)(2)-(3)(0)] - \bar j[(0)(2)-(3)(1)]+\bar k[(0)(0)-(-2)(1)] = \bar i(-4) - \bar j(-3)+\bar k(2) = (-4,3,2)$ . …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$O(0,0,0)$
›Reveal solutionSolution
The angle between a line and a plane is found from $\sin\phi=\dfrac{|\vec n\cdot\vec d|}{|\vec n||\vec d|}$ , which evaluates to $\sin^{-1}\!\left(\dfrac{6\sqrt2}{5\sqrt7}\right)$ .

Concept and Intuition

The angle between a line and a plane is the complement of the angle between the line and the plane's normal. If $\psi$ is the angle between the line's direction $\vec d$ and the normal $\vec n$ , then the line-plane angle is $\phi=90^\circ-\psi$ , so $\sin\phi=\cos\psi=\dfrac{|\vec n\cdot\vec d|}{|\vec n||\vec d|}$ — a sine formula, not a cosine formula, which is the key distinguishing feature from the line-normal or plane-plane angle formulas.

Step-by-Step Solution

Face $OAB$ contains $O,A(1,2,1),B(2,1,3)$ ; its normal is $\vec n=\vec{OA}\times\vec{OB}$ .

$\vec n=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&2&1\\2&1&3\end{vmatrix}=\mathbf i(6-1)-\mathbf j(3-2)+\mathbf k(1-4)=(5,-1,-3)$ .

Edge direction $\vec d=\vec{BC}=C-B=(-1-2,\,1-1,\,2-3)=(-3,0,-1)$ .

$\vec n\cdot\vec d=5(-3)+(-1)(0)+(-3)(-1)=-15+0+3=-12$ .

$|\vec n|=\sqrt{25+1+9}=\sqrt{35}$ , $|\vec d|=\sqrt{9+0+1}=\sqrt{10}$ .

$\sin\phi=\dfrac{12}{\sqrt{35}\sqrt{10}}=\dfrac{12}{\sqrt{350}}=\dfrac{12}{5\sqrt{14}}$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\dfrac{\pi}{2}$
›Reveal solutionSolution
Placing a cube vertex at the origin and writing down the coterminous space diagonal and face diagonal as vectors, the dot-product formula gives $\cos\theta=\sqrt{2/3}$ .

Concept and Intuition

"Coterminus" here means both diagonals start from the same vertex. Vector geometry makes 3D angle problems in a cube routine: assign coordinates to the cube's vertices, write each diagonal as a displacement vector from the shared vertex, and use $\cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$ .

Step-by-Step Solution

Take a unit cube with one vertex at the origin $O=(0,0,0)$ and edges along the axes, so the opposite vertex is $(1,1,1)$ .

The space (body) diagonal from $O$ is the vector $\vec d_1=(1,1,1)$ , with $|\vec d_1|=\sqrt{3}$ .

A face diagonal from the same vertex $O$ , lying in the face $z=0$ , goes to $(1,1,0)$ : $\vec d_2=(1,1,0)$ , with $|\vec d_2|=\sqrt{2}$ .

Dot product: $\vec d_1\cdot\vec d_2=1(1)+1(1)+1(0)=2$ .

$\cos\theta=\dfrac{2}{\sqrt3\cdot\sqrt2}=\dfrac{2}{\sqrt6}=\dfrac{2}{\sqrt6}\cdot\dfrac{\sqrt6}{\sqrt6}=\dfrac{2\sqrt6}{6}=\dfrac{\sqrt6}{3}$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\angle BAC =$
›Reveal solutionSolution
Form the two vectors from $A$ and dot them — the dot product vanishes, so the angle is a right angle. Answer: $90°$ .

Concept and Intuition

The angle at vertex $A$ between rays $AB$ and $AC$ is found from $\cos(\angle BAC)=\dfrac{\vec{AB}\cdot\vec{AC}}{|\vec{AB}||\vec{AC}|}$ . If the numerator (the dot product) is zero, the angle is exactly $90°$ regardless of the vector magnitudes — so it's worth checking the dot product first before computing any magnitudes.

Step-by-Step Solution

$\vec{AB}=B-A=(5-0,\,0-4,\,12-(-3))=(5,-4,15)$ .

$\vec{AC}=C-A=(7-0,\,24-4,\,0-(-3))=(7,20,3)$ .

Dot product: $\vec{AB}\cdot\vec{AC}=(5)(7)+(-4)(20)+(15)(3)=35-80+45=0$ . …
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AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$c$
›Reveal solutionSolution
This tests translating "angle is obtuse for all $x$ " into "dot product is negative for all $x$ " and then solving the resulting quadratic-in- $x$ inequality for the parameter $c$ . The answer is $c\in\left(-\tfrac43,0\right]$ .

Concept and Intuition

The angle between two vectors is obtuse exactly when their dot product is negative (and they aren't anti-parallel making it exactly $180^\circ$ , which doesn't arise as a boundary issue here). So "obtuse for all real $x$ " becomes: the expression $\bar a\cdot\bar b$ , viewed as a quadratic in $x$ with $c$ as parameter, is negative for every $x$ . That requires the parabola (in $x$ ) to open downward with no real roots — or be a negative constant.

Step-by-Step Solution

$\bar a\cdot\bar b=(cx)(x)+(-6)(2)+(3)(2cx)=cx^2-12+6cx=cx^2+6cx-12$ .

Need $f(x)=cx^2+6cx-12<0$ for all real $x$ .

Case $c=0$ : $f(x)=-12<0$ always — satisfies the condition.

Case $c\ne0$ : for a quadratic to be negative for all $x$ , it must open downward ( $c<0$ ) and have no real roots (discriminant $<0$ ). …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$\bar{f}, \bar{g}, \bar{h}$
›Reveal solutionSolution
Uses orthogonality to kill cross dot products; answer is $\cos^{-1}(1/\sqrt3)$ .

Concept and Intuition

When three vectors are mutually perpendicular and of equal magnitude, their sum is the space-diagonal of a cube built on them. The angle any diagonal makes with an edge is a classic $\cos^{-1}(1/\sqrt3)$ result.

Step-by-Step Solution

Let $|\bar f|=|\bar g|=|\bar h|=a$ , and $\bar f\cdot\bar g=\bar g\cdot\bar h=\bar h\cdot\bar f=0$ .

$(\bar f+\bar g+\bar h)\cdot\bar h = \bar f\cdot\bar h + \bar g\cdot\bar h + \bar h\cdot\bar h = 0+0+a^2 = a^2$ .

$|\bar f+\bar g+\bar h|^2 = |\bar f|^2+|\bar g|^2+|\bar h|^2 + 2(\bar f\cdot\bar g+\bar g\cdot\bar h+\bar h\cdot\bar f) = 3a^2$ , so $|\bar f+\bar g+\bar h|=a\sqrt3$ . …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$\bar{a}, \bar{b}$
›Reveal solutionSolution
Expand the perpendicularity condition $\bar c\cdot\bar d=0$ to isolate $\bar a\cdot\bar b$ .

Concept and Intuition

Two vectors are perpendicular exactly when their dot product is zero. Expanding the dot product of linear combinations of unit vectors reduces everything to the single unknown $\bar a\cdot\bar b=\cos\theta$ .

Step-by-Step Solution

$\bar c\cdot\bar d = (\bar a+2\bar b)\cdot(5\bar a-4\bar b) = 5(\bar a\cdot\bar a) -4(\bar a\cdot\bar b) + 10(\bar b\cdot\bar a) -8(\bar b\cdot\bar b)$ .

Since $|\bar a|=|\bar b|=1$ : $=5(1) + 6(\bar a\cdot\bar b) - 8(1) = 6(\bar a\cdot\bar b) - 3$ .

Set to zero: $6(\bar a\cdot\bar b)=3\Rightarrow \bar a\cdot\bar b = \dfrac12$ . …
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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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