(2a + 3b) (5a + 7b) = - Vidyadip

MCQ

$(2a + 3b) \times (5a + 7b) = $

A
$a \times b$
✓
$b \times a$
C
$a + b$
D
$7a + 10b$

✓

Answer

Correct option: B.

$b \times a$

b
(b) $14(a \times b) + 15(b \times a) = b \times a$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 10. vector algebra→10. vector algebra — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $\text{f(x)}=|\text{x}-\text{a}|\ \phi\ (\text{x}),$ where $\phi(\text{x})$ is continuous function, then:

$\text{f}'(\text{a}^+)=\phi(\text{a})$
$\text{f}'(\text{a}^-)=-\phi(\text{a})$
$\text{f}'(\text{a}^+)=\text{f}'(\text{a}^-)$
None of these

$\frac{\text{d}}{\text{dx}}\bigg[\log\bigg\{\text{e}^\text{x}\Big(\frac{\text{x}-2}{\text{x}+2}\Big)^\frac{3}{4}\bigg\}\bigg]$ equals:

$\frac{\text{x}^2-1}{\text{x}^2-4}$
$1$
$\frac{\text{x}^2+1}{\text{x}^2-4}$
$\text{e}^\text{x}\frac{\text{x}^2-1}{\text{x}^2-4}$

If $f(x) = x(\sqrt x - \sqrt {x + 1} ),$ then

The maximum value of the function ${x^3} + {x^2} + x - 4$ is

If the three points A(1, 6), B(3, −4) and C(x, y) are collinear, then the equation satisfying by x and y is:

5x + y − 11 = 0
5x + 13y + 5= 0
5x − 13y + 5 = 0
13x − 5y + 5 = 0

Number of real values of $\lambda$ for which the matrix $A =$ $\left[ {\begin{array}{*{20}{c}}{\lambda - 1}&\lambda &{\lambda + 1}\\2&{ - 1}&3\\ {\lambda + 3}&{\lambda - 2}&{\lambda + 7}\end{array}} \right]$ has no inverse

$\int\frac{1}{1-\cos\text{x}-\sin\text{x}}\text{ dx}=$

$\log\big|1+\cot\frac{\pi}{2}\big|+\text{C}$
$\log\big|1-\tan\frac{\pi}{2}\big|+\text{C}$
$\log\big|1-\cot\frac{\pi}{2}\big|+\text{C}$
$\log\big|1+\tan\frac{\pi}{2}\big|+\text{C}$

The value of the integral $\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$ is :

If $a + b + c = 0,\,\,\left| {\vec a} \right| = 3,\,\left| {\vec b} \right| = 5$ and $\left| {\vec c} \right| = 7,$ then the angle between $\vec a$ and $\vec b$ is

Let $P=\left[\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{array}\right]$, where $\alpha \in \mathbb{R}$. Suppose $Q=\left[q_{i j}\right]$ is a matrix such that $P Q=k I$, where $k \in \mathbb{R}, k \neq 0$ and $I$ is the identity matrix of order $3$ . If $q_{23}=-\frac{k}{8}$ and $\operatorname{det}(Q)=\frac{k^2}{2}$, then

($A$) $\quad \alpha=0, k=8$

($B$) $4 \alpha-k+8=0$

($C$) $\operatorname{det}(P \operatorname{adj}(Q))=2^9$

($D$) $\operatorname{det}(Q \operatorname{adj}(P))=2^{13}$