If {a} = {i} + {j} + {k}, {c} = {j} - {k}, {a} × {b} = {c} and {a} · {b} = 3 then {b} =

Correct Answer is : 13(5i‾+2j‾+2k‾)\; 1/3 (5 {i} + 2 {j} + 2 {k})31(5i+2j+2k)

Correct Answer is : 13(5i‾+2j‾+2k‾)\; 1/3 (5 {i} + 2 {j} + 2 {k})31(5i+2j+2k)

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AP EAPCET 06-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 32 of 160

Marks: +1, -0

\overline{a} = \overline{i} + \overline{j} + \overline{k}, \overline{c} = \overline{j} - \overline{k}, \overline{a} \times \overline{b} = \overline{c}

\overline{a} \cdot \overline{b} = 3

\overline{b} =

Solution: 👈: Video Solution

Let

\mathbf{b} = x \hat{i} + y \hat{j} + z \hat{k}

\mathbf{a} \times \mathbf{b} = \mathbf{c}

, then

\mathbf{c}

is perpendicular to both

\mathbf{a}

and

\mathbf{b}

\mathbf{b} \cdot \mathbf{c} = 0

y - z = 0

y = z

......(i)

Also a

\cdot \mathbf{b} = 3

x + y + z = 3

x + 2y = 3

.......(ii)

\; \mathbf{a} \times \mathbf{b} = \mathbf{c}

\; \;\; \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{j} - \hat{k}

\; \hat{i}(z - y) - \hat{j}(z - x) + \hat{k}(y - x) = \hat{j} - \hat{k}

\;\Rightarrow \;\; x - z = 1 \; \text{ and } \; x - y = 1

......(iii)

Subtracting Eq. (ii) from Eq. (i),

3y = 2 \Rightarrow y = \frac{2}{3}

Also,

z = \frac{2}{3}

[from Eq. (i)]

\; x + y + z = 3

\Rightarrow \;\; x + \frac{4}{3} = 3 \Rightarrow x = \frac{5}{3}

\therefore \;\; \mathbf{b} = \frac{1}{3} (5 \hat{i} + 2 \hat{j} + 2 \hat{k})

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