text{If } vec{a} = hat{i} + hat{j} + hat{k}, | vedclass

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

Question

(B) Given: $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}, \vec{c} = \hat{i} - \hat{j}$.First,calculate the cross pro

Vedclass · Accepted Answer

$(\frac{4}{5}, \frac{11}{5}, \frac{19}{5})$

Vedclass · Answer

(B) Given: $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}, \vec{c} = \hat{i} - \hat{j}$.First,calculate the cross products:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & -1 & 3 \end{vmatrix} = \hat{i}(3+1) - \hat{j}(3-2) + \hat{k}(-1-2) = 4\hat{i} - \hat{j} - 3\hat{k}$.$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 3 \ 1 & -1 & 0 \end{vmatrix} = \hat{i}(0+3) - \hat{j}(0-3) + \hat{k}(-2+1) = 3\hat{i} + 3\hat{j} - \hat{k}$.$\vec{c} 	imes \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 0 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-1-0) - \hat{j}(1-0) + \hat{k}(1+1) = -\hat{i} - \hat{j} + 2\hat{k}$.Now,equate $6\hat{i} + 2\hat{j} + 3\hat{k} = \lambda_1(4\hat{i} - \hat{j} - 3\hat{k}) + \lambda_2(3\hat{i} + 3\hat{j} - \hat{k}) + \lambda_3(-\hat{i} - \hat{j} + 2\hat{k})$.Comparing coefficients of $\hat{i}, \hat{j}, \hat{k}$:$4\lambda_1 + 3\lambda_2 - \lambda_3 = 6$ $(i)$$-\lambda_1 + 3\lambda_2 - \lambda_3 = 2$ (ii)$-3\lambda_1 - \lambda_2 + 2\lambda_3 = 3$ (iii)Subtracting (ii) from $(i)$: $5\lambda_1 = 4 \Rightarrow \lambda_1 = \frac{4}{5}$.Substitute $\lambda_1$ into (ii) and (iii):$3\lambda_2 - \lambda_3 = 2 + \frac{4}{5} = \frac{14}{5}$ (iv)$-\lambda_2 + 2\lambda_3 = 3 + 3(\frac{4}{5}) = \frac{27}{5}$ $(v)$Multiply (iv) by $2$ and add to $(v)$: $6\lambda_2 - 2\lambda_3 + (-\lambda_2 + 2\lambda_3) = \frac{28}{5} + \frac{27}{5} \Rightarrow 5\lambda_2 = \frac{55}{5} = 11 \Rightarrow \lambda_2 = \frac{11}{5}$.From (iv): $\lambda_3 = 3(\frac{11}{5}) - \frac{14}{5} = \frac{33-14}{5} = \frac{19}{5}$.Thus,$(\lambda_1, \lambda_2, \lambda_3) = (\frac{4}{5}, \frac{11}{5}, \frac{19}{5})$.

Similar Questions

$A$ unit vector which is perpendicular to the vector $2\hat{i} - \hat{j} + 2\hat{k}$ and is coplanar with the vectors $\hat{i} + \hat{j} - \hat{k}$ and $2\hat{i} + 2\hat{j} - \hat{k}$ is

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors. Suppose $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{6}$. Then $\vec{a}$ is

Let vectors $a, b, c$ and $d$ be such that $(a \times b) \times (c \times d) = 0$. If $a$ and $b$ lie in plane $P_1$ and $c$ and $d$ lie in plane $P_2$,find the angle between $P_1$ and $P_2$.

If $A, B, C, D$ are four points in space,then $|\overline{AB} \times \overline{CD} + \overline{BC} \times \overline{AD} + \overline{CA} \times \overline{BD}| = \lambda \times (\text{Area of } \Delta ABC)$. Find $\lambda$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module