Prove by vector method that $\cos(A - B) = \cos A \cos B + \sin A \sin B$.

> [!TLDR] > Represent A and B by unit vectors at those angles, compute their dot product two ways (geometric definition, and by components), and equate. > > 1. In the $xy$-plane, let $\hat p=\overrightarrow{OP}$ and $\hat q=\overrightarrow{OQ}$ be unit vectors making angles $A$ and $B$ respectively with the positive $x$-axis (measuring angles in the standard sense). > 2. In component form: $\hat p=\cos A\,\vec i+\sin A\,\vec j$ and $\hat q=\cos B\,\vec i+\sin B\,\vec j$, and $|\hat p|=|\hat q|=1$. > 3. The angle between the directions $\hat p$ and $\hat q$ is $(A-B)$ (or $(B-A)$; since cosine is an even function, $\cos(A-B)=\cos(B-A)$, so the sign convention does not matter). > 4. By the geometric definition of the dot product: $\hat p\cdot\hat q=|\hat p||\hat q|\cos(A-B)=(1)(1)\cos(A-B)=\cos(A-B)$. …

Prove by vector method that cos(A - B) = cos A cos – Class 12 Mathematics

Question 127 of 162

\cos(A - B) = \cos A \cos B + \sin A \sin B

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018Subjective· 10mImportance★★★★★

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Concept understanding — Applications of Dot and Cross Product

The dot product $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ and cross product $\vec a\times\vec b$ (magnitude $|\vec a||\vec b|\sin\theta$ , direction perpendicular to both) are not just formulas — placed carefully, they reprove classical geometry theorems and compute two real mechanical quantities.

Vector proofs in geometry and trigonometry. The standard technique: put a convenient point (a triangle's vertex, a circle's centre) at the origin, write every other point as a position vector, and express every needed segment as a difference of position vectors. Then:

Perpendicularity is proved by showing a dot product is $0$ . E.g. if $O$ is a circle's centre and $M$ is the midpoint of chord $AB$ (position vectors $\vec a,\vec b$ with $|\vec a|=|\vec b|=r$ ), then $\vec{OM}=\tfrac{\vec a+\vec b}2$ and $\vec{AB}=\vec b-\vec a$ , and $\vec{OM}\cdot\vec{AB}=\tfrac12(|\vec b|^2-|\vec a|^2)=0$ .
Equal lengths / rectangles come from expanding $|\vec p\pm\vec q|^2=|\vec p|^2\pm2\vec p\cdot\vec q+|\vec q|^2$ and comparing.
Areas come from $\text{area of a parallelogram}=|\vec p\times\vec q|$ for adjacent sides $\vec p,\vec q$ ; a triangle is half that, and a general quadrilateral with diagonals $\vec d_1,\vec d_2$ has area $\tfrac12|\vec d_1\times\vec d_2|$ (proved by splitting along one diagonal, since the two triangles on either side add — same-sense cross products — instead of subtracting).
Compound-angle identities ( $\cos(\alpha\mp\beta)$ , $\sin(\alpha\pm\beta)$ ) drop out of dotting or crossing two unit vectors $\hat i\cos\alpha+\hat j\sin\alpha$ and $\hat i\cos\beta\pm\hat j\sin\beta$ placed at angles $\alpha,\pm\beta$ to the $x$ -axis. …

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Represent A and B by unit vectors at those angles, compute their dot product two ways (geometric definition, and by components), and equate.

In the $xy$ -plane, let $\hat p=\overrightarrow{OP}$ and $\hat q=\overrightarrow{OQ}$ be unit vectors making angles $A$ and $B$ respectively with the positive $x$ -axis (measuring angles in the standard sense).
In component form: $\hat p=\cos A\,\vec i+\sin A\,\vec j$ and $\hat q=\cos B\,\vec i+\sin B\,\vec j$ , and $|\hat p|=|\hat q|=1$ .
The angle between the directions $\hat p$ and $\hat q$ is $(A-B)$ (or $(B-A)$ ; since cosine is an even function, $\cos(A-B)=\cos(B-A)$ , so the sign convention does not matter).
By the geometric definition of the dot product: $\hat p\cdot\hat q=|\hat p||\hat q|\cos(A-B)=(1)(1)\cos(A-B)=\cos(A-B)$ . …

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

CBSE 2019Set ANNUAL1 markMCQ
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$
(a) $\dfrac{5\pi}{3}$
(b) $\dfrac{\pi}{2}$
(c) $\dfrac{\pi}{6}$
(d) $\dfrac{2\pi}{3}$
›Reveal solutionSolution
From $\vec a+\vec b+\vec c=\vec 0$ with $|\vec a|=3,|\vec b|=4,|\vec c|=5$ , the angle between $\vec a$ and $\vec b$ works out to $\dfrac{\pi}{2}$ .

From $\vec a+\vec b+\vec c=\vec0$ , we get $\vec c=-(\vec a+\vec b)$ .

Take magnitudes squared: $|\vec c|^2=|\vec a+\vec b|^2=|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2$ .

Substitute the given magnitudes: $5^2=3^2+2\vec a\cdot\vec b+4^2$ , i.e. $25=9+16+2\vec a\cdot\vec b=25+2\vec a\cdot\vec b$ .

So $2\vec a\cdot\vec b=0\Rightarrow \vec a\cdot\vec b=0$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$\vec{PR} = 2\vec i + \vec j + \vec k$
(a) $5\sqrt3$
(b) $10\sqrt3$
(c) $\dfrac{5\sqrt3}{2}$
(d) $\dfrac{3}{2}$
›Reveal solutionSolution
The area of a quadrilateral in terms of its diagonal vectors $\vec d_1,\vec d_2$ is $\frac12|\vec d_1\times\vec d_2|$ ; here the diagonals are $\vec{PR}$ and $\vec{QS}$ , so compute their cross product and halve its magnitude.

For quadrilateral PQRS, the diagonals are $PR$ and $QS$ , and the standard vector-algebra result is $\text{Area}(PQRS) = \frac12\,|\vec{PR}\times\vec{QS}|.$

Given $\vec{PR}=2\vec i+\vec j+\vec k$ and $\vec{QS}=-\vec i+3\vec j+2\vec k$ .

Compute the cross product: $\vec{PR}\times\vec{QS}=\begin{vmatrix}\vec i & \vec j & \vec k\\ 2 & 1 & 1\\ -1 & 3 & 2\end{vmatrix}$

$\vec i$ -component: $(1)(2)-(1)(3) = 2-3=-1$ .

$\vec j$ -component: $-[(2)(2)-(1)(-1)] = -(4+1)=-5$ . …
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CBSE 2016Set ANNUAL1 markMCQ
$\vec{i} - \vec{j}$
(a) $0$
(b) $1$
(c) $-1$
(d) $2$
›Reveal solutionSolution
The vector $\vec i-\vec j$ has zero component along $\vec k$ , so its projection on the $z$ -axis is $0$ .

The projection of a vector $\vec r=x\vec i+y\vec j+z\vec k$ on the $z$ -axis equals $\vec r\cdot\vec k=z$ .

Here $\vec r=\vec i-\vec j=1\vec i-1\vec j+0\vec k$ .

So $\vec r\cdot\vec k=0$ . …
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CBSE 2016Set ANNUAL1 markMCQ
$3\vec i + \vec j - \vec k$
(a) $10\sqrt3$
(b) $6\sqrt{30}$
(c) $\dfrac{3}{2}\sqrt{30}$
(d) $3\sqrt{30}$
›Reveal solutionSolution
Recovering the second side from the diagonal and given side, then taking the cross product, gives an area of $3\sqrt{30}$ .

In a parallelogram with adjacent sides $\vec a$ and $\vec b$ from the same vertex, the diagonal from that vertex is $\vec d=\vec a+\vec b$ .

Given side $\vec a=\vec i-3\vec j+4\vec k$ and diagonal $\vec d=3\vec i+\vec j-\vec k$ , the other side is $\vec b=\vec d-\vec a=(3-1)\vec i+(1-(-3))\vec j+(-1-4)\vec k=2\vec i+4\vec j-5\vec k$ .

Area of the parallelogram $=|\vec a\times\vec b|$ .

Compute the cross product: $\vec a\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\1&-3&4\\2&4&-5\end{vmatrix}$ $i$ -component: $(-3)(-5)-(4)(4)=15-16=-1$ . $j$ -component: $-[(1)(-5)-(4)(2)]=-[-5-8]=13$ . …
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Q.Prove by vector method that cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin Bcos(A−B)=cosAcosB+sinAsinB.

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Q.Prove by vector method that $\cos(A - B) = \cos A \cos B + \sin A \sin B$ .

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