The vector b¯=3i¯+4j¯ is to be written as the sum of a vector b¯1, parallel to a¯=i¯+j¯ and a vector b¯2 perpendicular to a¯ then b1¯= _______

b¯1|| a¯ ⇒ b¯1=c(i¯+j¯) b¯2=b¯−b¯1=(3−c)i¯−cj¯+4k¯ Also b¯2⊥a¯ ⇒ b¯2.a¯=0 ⇒(3−c)−c=0 ⇒c=32 Hence b¯1=32(i¯+j¯) .

The vector b¯=3i¯+4j¯ is to be written as the sum of a vector b¯1, parallel to a¯=i¯+j¯ and a vector b¯2 perpendicular to a¯ then b1¯= _______

b¯1|| a¯ ⇒ b¯1=c(i¯+j¯) b¯2=b¯−b¯1=(3−c)i¯−cj¯+4k¯ Also b¯2⊥a¯ ⇒ b¯2.a¯=0 ⇒(3−c)−c=0 ⇒c=32 Hence b¯1=32(i¯+j¯) .

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

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

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

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

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

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Detailed Solution

${\bar{b}}_{1} | | \bar{a} ⇒ {\bar{b}}_{1} = c (\bar{i} + \bar{j}) {\bar{b}}_{2} = \bar{b} - {\bar{b}}_{1} = (3 - c) \bar{i} - c \bar{j} + 4 \bar{k} A l s o {\bar{b}}_{2} ⊥ \bar{a} \Rightarrow {\bar{b}}_{2} . \bar{a} = 0 \Rightarrow (3 - c) - c = 0 \Rightarrow c = \frac{3}{2} H e n c e {\bar{b}}_{1} = \frac{3}{2} (\bar{i} + \bar{j}) .$