If vec{a}=hat{i}+hat{j}+hat{k} and vec{b}=2ha | vedclass

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

Question

(D) Any vector $\vec{c}$ lying in the plane of $\vec{a}$ and $\vec{b}$ can be written as $\vec{c} = x\vec{a} + y\vec{b}$.Since $\vec{c}$ is perpendicu

Vedclass · Accepted Answer

$\frac{2}{\sqrt{21}}$

Vedclass · Answer

(D) Any vector $\vec{c}$ lying in the plane of $\vec{a}$ and $\vec{b}$ can be written as $\vec{c} = x\vec{a} + y\vec{b}$.Since $\vec{c}$ is perpendicular to $\vec{b}$,we have $\vec{c} \cdot \vec{b} = 0$.Substituting $\vec{c} = x\vec{a} + y\vec{b}$,we get $(x\vec{a} + y\vec{b}) \cdot \vec{b} = 0$,which implies $x(\vec{a} \cdot \vec{b}) + y|\vec{b}|^2 = 0$.Given $\vec{a} \cdot \vec{b} = (1)(2) + (1)(1) + (1)(1) = 4$ and $|\vec{b}|^2 = 2^2 + 1^2 + 1^2 = 6$.So,$4x + 6y = 0 \Rightarrow 2x + 3y = 0$. Let $x = 3k$ and $y = -2k$.Then $\vec{c} = 3k(\hat{i}+\hat{j}+\hat{k}) - 2k(2\hat{i}+\hat{j}+\hat{k}) = k(-\hat{i} + \hat{j} + \hat{k})$.Since $\vec{c}$ is a unit vector,$|\vec{c}| = |k|\sqrt{(-1)^2 + 1^2 + 1^2} = |k|\sqrt{3} = 1 \Rightarrow k = \frac{1}{\sqrt{3}}$.Thus,$\vec{c} = \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$.Finally,$\vec{c} \cdot (\hat{i}+\hat{j}+2\hat{k}) = \frac{1}{\sqrt{3}}(-1 + 1 + 2) = \frac{2}{\sqrt{3}}$.Wait,re-evaluating the plane condition: $\vec{c}$ is in the plane of $\vec{a}$ and $\vec{b}$ and $\vec{c} \perp \vec{b}$.Let $\vec{n} = \vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & 1 & 1 \end{vmatrix} = 0\hat{i} + 1\hat{j} - 1\hat{k} = \hat{j} - \hat{k}$.Then $\vec{c}$ must be parallel to $\vec{n} 	imes \vec{b} = (\hat{j} - \hat{k}) 	imes (2\hat{i} + \hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -1 \ 2 & 1 & 1 \end{vmatrix} = 2\hat{i} - 2\hat{j} - 2\hat{k}$.Unit vector $\vec{c} = \pm \frac{2\hat{i} - 2\hat{j} - 2\hat{k}}{\sqrt{4+4+4}} = \pm \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} - \hat{k})$.Calculating $\vec{c} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = \pm \frac{1}{\sqrt{3}}(1 - 1 - 2) = \mp \frac{2}{\sqrt{3}}$.

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+\hat{k}$ are two vectors,and $\vec{c}$ is a unit vector lying in the plane of $\vec{a}$ and $\vec{b}$ such that $\vec{c}$ is perpendicular to $\vec{b}$,then find the value of $\vec{c} \cdot (\hat{i}+\hat{j}+2\hat{k})$.

Similar Questions

$A$ unit vector perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ is:

Let $|\vec{a}|=2, |\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Then $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$ is equal to

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two perpendicular lines,then the direction cosines of the line which is perpendicular to both the lines will be:

If $a=2\hat{i}+\hat{j}-3\hat{k}$,$b=\hat{i}-2\hat{j}+\hat{k}$,$c=-\hat{i}+\hat{j}-4\hat{k}$ and $d=\hat{i}+\hat{j}+\hat{k}$,then $|(a \times b) \times(c \times d)|=$

The adjacent sides of a parallelogram are $\bar{a} = \hat{j} + 2\hat{k}$ and $\bar{b} = \hat{i} + 2\hat{j}$. Find its area.

Vedclass Test Series

Exam Paper Generator

Online Exam Module