If overline{r} = -4 hat{i} - 6 hat{j} - 2 hat | vedclass

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

Question

(A) Let $\overline{r} = x \overline{a} + y \overline{b}$.Substituting the given vectors:$-4 \hat{i} - 6 \hat{j} - 2 \hat{k} = x(-\hat{i} - 4 \hat{j} +

Vedclass · Accepted Answer

$\overline{r} = -\frac{4}{3} \overline{a} + \frac{2}{3} \overline{b}$

Vedclass · Answer

(A) Let $\overline{r} = x \overline{a} + y \overline{b}$.Substituting the given vectors:$-4 \hat{i} - 6 \hat{j} - 2 \hat{k} = x(-\hat{i} - 4 \hat{j} + 3 \hat{k}) + y(-8 \hat{i} - \hat{j} + 3 \hat{k})$$= (-x - 8y) \hat{i} + (-4x - y) \hat{j} + (3x + 3y) \hat{k}$Comparing the components,we get:$(1)$ $-x - 8y = -4$$(2)$ $-4x - y = -6$$(3)$ $3x + 3y = -2$From $(3)$,$x + y = -\frac{2}{3}$,so $y = -\frac{2}{3} - x$.Substitute $y$ into $(1)$: $-x - 8(-\frac{2}{3} - x) = -4 \implies -x + \frac{16}{3} + 8x = -4 \implies 7x = -4 - \frac{16}{3} = -\frac{28}{3} \implies x = -\frac{4}{3}$.Then $y = -\frac{2}{3} - (-\frac{4}{3}) = \frac{2}{3}$.Thus,$\overline{r} = -\frac{4}{3} \overline{a} + \frac{2}{3} \overline{b}$.

If $\overline{r} = -4 \hat{i} - 6 \hat{j} - 2 \hat{k}$ is a linear combination of the vectors $\overline{a} = -\hat{i} - 4 \hat{j} + 3 \hat{k}$ and $\overline{b} = -8 \hat{i} - \hat{j} + 3 \hat{k}$,then which of the following is true?

Similar Questions

If $O$ is the origin and $C$ is the midpoint of $A(2, -1)$ and $B(-4, 3)$,then the value of $\overrightarrow{OC}$ is:

If $ABCDEF$ is a regular hexagon,where two adjacent sides $\vec{AB}$ and $\vec{BC}$ are $\vec{a}$ and $\vec{b}$ respectively. Then $\vec{CD}$ is

What should be added to the vector $a = 3i + 4j - 2k$ to obtain the resultant vector $i$?

The position vectors of $A$ and $B$ are $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 3\hat{k}$. The position vector of the midpoint of the line segment $AB$ is:

If the sum of two unit vectors is a unit vector,then the magnitude of their difference is

Vedclass Test Series

Exam Paper Generator

Online Exam Module