Let lambda in mathbb{Z},vec{a} = lambda hat{i | vedclass

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

Question

(B) Given $(\vec{a} + \vec{b} + \vec{c}) 	imes \vec{c} = \vec{0}$.Since $\vec{c} 	imes \vec{c} = \vec{0}$,this simplifies to $(\vec{a} + \vec{b})

Vedclass · Accepted Answer

$46$

Vedclass · Answer

(B) Given $(\vec{a} + \vec{b} + \vec{c}) 	imes \vec{c} = \vec{0}$.Since $\vec{c} 	imes \vec{c} = \vec{0}$,this simplifies to $(\vec{a} + \vec{b}) 	imes \vec{c} = \vec{0}$.This implies $\vec{c}$ is parallel to $(\vec{a} + \vec{b})$. Let $\vec{c} = \alpha(\vec{a} + \vec{b})$ for some scalar $\alpha$.$\vec{a} + \vec{b} = (\lambda + 3)\hat{i} + 0\hat{j} + 1\hat{k}$.So,$\vec{c} = \alpha(\lambda + 3)\hat{i} + \alpha\hat{k}$.Given $\vec{b} \cdot \vec{c} = -20 \Rightarrow (3\hat{i} - \hat{j} + 2\hat{k}) \cdot (\alpha(\lambda + 3)\hat{i} + \alpha\hat{k}) = -20$.$3\alpha(\lambda + 3) + 2\alpha = -20 \Rightarrow \alpha(3\lambda + 11) = -20$.Given $\vec{a} \cdot \vec{c} = -17 \Rightarrow (\lambda\hat{i} + \hat{j} - \hat{k}) \cdot (\alpha(\lambda + 3)\hat{i} + \alpha\hat{k}) = -17$.$\alpha\lambda(\lambda + 3) - \alpha = -17 \Rightarrow \alpha(\lambda^2 + 3\lambda - 1) = -17$.Dividing the two equations: $\frac{3\lambda + 11}{\lambda^2 + 3\lambda - 1} = \frac{20}{17}$.$17(3\lambda + 11) = 20(\lambda^2 + 3\lambda - 1) \Rightarrow 51\lambda + 187 = 20\lambda^2 + 60\lambda - 20$.$20\lambda^2 + 9\lambda - 207 = 0$. Solving for $\lambda \in \mathbb{Z}$,we get $\lambda = 3$.Substituting $\lambda = 3$ into $\alpha(3(3) + 11) = -20 \Rightarrow 20\alpha = -20 \Rightarrow \alpha = -1$.Thus,$\vec{c} = -1(6\hat{i} + \hat{k}) = -6\hat{i} - \hat{k}$.We need to find $|\vec{c} 	imes (3\hat{i} + \hat{j} + \hat{k})|^2$.Let $\vec{v} = \vec{c} 	imes (3\hat{i} + \hat{j} + \hat{k}) = (-6\hat{i} - \hat{k}) 	imes (3\hat{i} + \hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -6 & 0 & -1 \ 3 & 1 & 1 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(-6 - (-3)) + \hat{k}(-6 - 0) = \hat{i} + 3\hat{j} - 6\hat{k}$.$|\vec{v}|^2 = 1^2 + 3^2 + (-6)^2 = 1 + 9 + 36 = 46$.

Similar Questions

If the position vectors of the vertices of a triangle are $2\hat{i} - \hat{j} + \hat{k}$,$\hat{i} - 3\hat{j} - 5\hat{k}$,and $3\hat{i} - 4\hat{j} - 4\hat{k}$,then the triangle is:

If $\vec{a}, \vec{b}$ and $\vec{c}$ are mutually perpendicular vectors of equal magnitudes,show that the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.

$A$ unit vector in the $xy$-plane that makes an angle of $45^{\circ}$ with the vector $(i + j)$ and an angle of $60^{\circ}$ with the vector $(3i - 4j)$ is:

If $\overline{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}$,$\overline{b}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overline{c}=\hat{i}+3 \hat{j}$ are such that $(\overline{a}+\lambda \overline{b})$ is perpendicular to $\overline{c}$,then the value of $\lambda$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module