a, b, c are three vectors such that |a|=3, |b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given that $|a|=3, |b|=5, |c|=7$. Since $a$ is perpendicular to $(b+c)$,we have $a \cdot (b+c) = 0 \implies a \cdot b + a \cdot c = 0$ ... $(i)$ S

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Given that $|a|=3, |b|=5, |c|=7$. Since $a$ is perpendicular to $(b+c)$,we have $a \cdot (b+c) = 0 \implies a \cdot b + a \cdot c = 0$ ... $(i)$ Since $b$ is perpendicular to $(c+a)$,we have $b \cdot (c+a) = 0 \implies b \cdot c + b \cdot a = 0$ ... $(ii)$ Since $c$ is perpendicular to $(a+b)$,we have $c \cdot (a+b) = 0 \implies c \cdot a + c \cdot b = 0$ ... $(iii)$ Adding equations $(i), (ii),$ and $(iii)$,we get: $2(a \cdot b + b \cdot c + c \cdot a) = 0 \implies a \cdot b + b \cdot c + c \cdot a = 0$. Now,we calculate $|a+b+c|^2$: $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$ $|a+b+c|^2 = (3)^2 + (5)^2 + (7)^2 + 2(0) = 9 + 25 + 49 = 83$. Finally,we evaluate the required expression: $\sqrt{(a+b+c)^2 - 2} = \sqrt{83 - 2} = \sqrt{81} = 9$.

$a, b, c$ are three vectors such that $|a|=3, |b|=5, |c|=7$. If $a, b, c$ are perpendicular to the vectors $b+c, c+a, a+b$ respectively,then $\sqrt{(a+b+c)^2-2}=$

Similar Questions

For vectors $\vec{a}, \vec{b}$ and $\vec{c}$,if $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=2, |\vec{b}|=3, |\vec{c}|=5$,then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is . . . . . . .

The orthogonal projection of vector $\vec{a}$ on vector $\vec{b}$ is:

If $\overrightarrow{F_1} = i - j + k,$ $\overrightarrow{F_2} = -i + 2j - k,$ $\overrightarrow{F_3} = j - k,$ $\vec{A} = 4i - 3j - 2k$ and $\vec{B} = 6i + j - 3k,$ then the scalar product of $(\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3})$ and $\overrightarrow{AB}$ will be:

If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b},$ then $\vec{a} \cdot \vec{b} \ge 0$ if

If the lines $\vec{r} = 2\hat{i} + \hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and $\vec{r} = \hat{i} + \hat{j} - 3\hat{k} + \mu(\hat{j} + 2\hat{k})$ intersect each other,then $(\lambda + \mu)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module