Discuss the properties of scalar and vector products.

III. Long Answer Questions · Q2

Q. Discuss the properties of scalar and vector products.

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✓ Free question

Step 1. Scalar (dot) product properties. $\vec A\cdot\vec B=AB\cos\theta$ ; always a scalar; positive for acute $\theta$ , negative for obtuse $\theta$ ; commutative ( $\vec A\cdot\vec B=\vec B\cdot\vec A$ ); distributive over addition; maximum ( $=AB$ ) when $\theta=0°$ (parallel); minimum ( $=-AB$ ) when $\theta=180°$ (anti-parallel); zero when $\theta=90°$ (perpendicular — the test for orthogonality); self-dot-product $\vec A\cdot\vec A=A^2$ ; for orthogonal unit vectors $\hat i\cdot\hat i=1$ , $\hat i\cdot\hat j=0$ (and cyclic); component form $\vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z$ .

Step 2. Vector (cross) product properties. $\vec A\times\vec B=(AB\sin\theta)\hat n$ , with $\hat n$ perpendicular to the plane of $\vec A,\vec B$ by the right-hand rule; always a vector; not commutative — $\vec A\times\vec B=-(\vec B\times\vec A)$ though both have the same magnitude $AB\sin\theta$ ; maximum magnitude ( $=AB$ ) when $\theta=90°$ (perpendicular); zero when $\theta=0°$ or $180°$ (parallel/anti-parallel, including the self-product $\vec A\times\vec A=0$ ); for orthogonal unit vectors $\hat i\times\hat j=\hat k$ , $\hat j\times\hat k=\hat i$ , $\hat k\times\hat i=\hat j$ (cyclic), with the reverse order giving the negative; component form via the $3\times3$ determinant recipe; $|\vec A\times\vec B|$ equals the area of the parallelogram with sides $\vec A,\vec B$ , and $\tfrac12|\vec A\times\vec B|$ the area of the triangle with sides $\vec A,\vec B$ .

Step 3. Contrast. The dot product measures how much two vectors point 'along' each other (peaking when parallel, vanishing when perpendicular); the cross product measures how much they point 'across' each other (peaking when perpendicular, vanishing when parallel) and, unlike the dot product, produces a brand-new vector rather than a number.

Step 4. Physical uses. Dot product: work $W=\vec F\cdot\vec d$ . Cross product: torque $\vec\tau=\vec r\times\vec F$ , angular momentum $\vec L=\vec r\times\vec p$ , and $\vec v=\vec\omega\times\vec r$ .

✓Final answer

AB\cos\theta

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

CBSE 2026Set ANNUAL2 marks
$p$
›Reveal solutionSolution

Perpendicular vectors have dot product zero: $\bar a\cdot\bar b=0$ .

$\bar a\cdot\bar b=(3)(1)+(2)(p)+(9)(3)=3+2p+27=2p+30$

For $\bar a\perp\bar b$ : $2p+30=0 \implies p=-15$ .

✓Final answer
$p=-15$
CBSE 2025Set ANNUAL2 marksMCQ
$|\bar a|=5$
(a) 30
(b) 60
(c) 40
(d) 45
›Reveal solutionSolution

Use $|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta$ to find $\theta$ , then $|\bar a\cdot\bar b|=|\bar a||\bar b|\cos\theta$ .

Given $|\bar a|=5$ , $|\bar b|=13$ , $|\bar a\times\bar b|=25$ .

$|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta \implies 25 = 5\cdot13\cdot\sin\theta = 65\sin\theta$

$\sin\theta=\frac{25}{65}=\frac{5}{13}$

Since $\sin\theta=5/13$ , $\cos\theta=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$ .

$|\bar a\cdot\bar b|=|\bar a||\bar b|\cos\theta = 5\cdot13\cdot\frac{12}{13}=5\cdot12=60$

✓Final answer
$|\bar a\cdot\bar b|=60$
CBSE 2023Set 1A2 marks
$\bar a$
›Reveal solutionSolution

$|\bar a+\bar b|=|\bar a-\bar b|$ forces $\bar a\cdot\bar b=0$ , so the vectors are perpendicular.

Step 1 — Square both sides.

$|\bar a+\bar b|^2 = |\bar a-\bar b|^2$

$(\bar a+\bar b)\cdot(\bar a+\bar b) = (\bar a-\bar b)\cdot(\bar a-\bar b)$

$|\bar a|^2+2\bar a\cdot\bar b+|\bar b|^2 = |\bar a|^2-2\bar a\cdot\bar b+|\bar b|^2$

Step 2 — Simplify.

$4\,\bar a\cdot\bar b = 0 \implies \bar a\cdot\bar b = 0$

Step 3 — Interpret.

Since $\bar a,\bar b$ are non-zero, $\bar a\cdot\bar b = |\bar a||\bar b|\cos\theta = 0$ forces $\cos\theta=0$ , i.e. $\theta = 90^\circ$ .

✓Final answer
$\bar a$
CBSE 2022Set 1A2 marks
$\lambda$
›Reveal solutionSolution

Two vectors are perpendicular exactly when their dot product is zero; solving that equation gives $\lambda$ .

Given $\bar{u} = \bar{i} - \lambda\bar{j} + 2\bar{k}$ and $\bar{v} = 8\bar{i} + 6\bar{j} - \bar{k}$ .

Step 1. Vectors are at right angles iff $\bar{u}\cdot\bar{v} = 0$ .

Step 2. Compute the dot product:

$\bar{u}\cdot\bar{v} = (1)(8) + (-\lambda)(6) + (2)(-1) = 8 - 6\lambda - 2 = 6 - 6\lambda$

Step 3. Set equal to zero: $6 - 6\lambda = 0 \Rightarrow \lambda = 1$ .

✓Final answer
$\lambda = 1$
CBSE 2019Set 1A2 marks
$\lambda\bar{i} - 3\bar{j} + 5\bar{k}$
›Reveal solutionSolution

Two vectors are perpendicular exactly when their dot product is zero — set the dot product to zero and solve the resulting quadratic in $\lambda$ .

Given $\bar{u}=\lambda\bar{i}-3\bar{j}+5\bar{k}$ and $\bar{v}=2\lambda\bar{i}-\lambda\bar{j}-\bar{k}$ , perpendicular means $\bar{u}\cdot\bar{v}=0$ .

$\bar{u}\cdot\bar{v} = \lambda(2\lambda) + (-3)(-\lambda) + 5(-1) = 2\lambda^2 + 3\lambda - 5 = 0$

Solve using the quadratic formula:

$\lambda = \dfrac{-3 \pm \sqrt{9+40}}{4} = \dfrac{-3\pm7}{4}$

$\lambda = \dfrac{4}{4}=1$ or $\lambda = \dfrac{-10}{4}=-\dfrac{5}{2}$

✓Final answer
$\lambda = 1$

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Q.Discuss the properties of scalar and vector products.

Concept understanding — Scalar (Dot) Product of Vectors

Q.Discuss the properties of scalar and vector products.

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