The vector b¯=3i¯+4k¯ is to be written as the sum of a vector b¯1 parallel a¯=i¯+j¯ and a vector b¯2 perpendicular to a¯ then b¯1 is equal to

Since b¯1∥a¯∴b¯1=c(i¯+j¯)b¯2=b¯=b¯1=(3−c)i¯−cj¯+4k¯Also b¯2⋅a¯=0⇒(3−c)−c=0⇒c=32Hence b¯1=32(i¯+j¯)

The vector b¯=3i¯+4k¯ is to be written as the sum of a vector b¯1 parallel a¯=i¯+j¯ and a vector b¯2 perpendicular to a¯ then b¯1 is equal to

Since b¯1∥a¯∴b¯1=c(i¯+j¯)b¯2=b¯=b¯1=(3−c)i¯−cj¯+4k¯Also b¯2⋅a¯=0⇒(3−c)−c=0⇒c=32Hence b¯1=32(i¯+j¯)

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The vector $\bar{b} = 3 \bar{i} + 4 \bar{k}$ is to be written as the sum of a vector ${\bar{b}}_{1}$ parallel $\bar{a} = \bar{i} + \bar{j}$ and a vector ${\bar{b}}_{2}$ perpendicular to $\bar{a}$ then ${\bar{b}}_{1}$ is equal to

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$\frac{1}{2} (\bar{i} + \bar{j})$

$\frac{3}{2} (\bar{i} + \bar{j})$

$\frac{2}{3} (\bar{i} + \bar{j})$

$\frac{1}{3} (\bar{i} + \bar{j})$

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Since ${\bar{b}}_{1} ∥ \bar{a}$

$∴ {\bar{b}}_{1} = c (\bar{i} + \bar{j})$

${\bar{b}}_{2} = \bar{b} = {\bar{b}}_{1} = (3 - c) \bar{i} - c \bar{j} + 4 \bar{k}$

Also ${\bar{b}}_{2} \cdot \bar{a} = 0 \Rightarrow (3 - c) - c = 0 \Rightarrow c = \frac{3}{2}$

Hence ${\bar{b}}_{1} = \frac{3}{2} (\bar{i} + \bar{j})$

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The vector b¯=3i¯+4k¯ is to be written as the sum of a vector b¯1 parallel a¯=i¯+j¯ and a vector b¯2 perpendicular to a¯ then b¯1 is equal to

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The vector $\bar{b} = 3 \bar{i} + 4 \bar{k}$ is to be written as the sum of a vector ${\bar{b}}_{1}$ parallel $\bar{a} = \bar{i} + \bar{j}$ and a vector ${\bar{b}}_{2}$ perpendicular to $\bar{a}$ then ${\bar{b}}_{1}$ is equal to