Let a=hat{i}+hat{j}, b=hat{j}+hat{k} and c=ha | vedclass

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(B) Let $d = d_1 \hat{i} + d_2 \hat{j} + d_3 \hat{k}$.Since $d$ is a unit vector, $d_1^2 + d_2^2 + d_3^2 = 1$ $(i)$.Given $a \cdot d = 0$, where $a =

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$\pm \frac{1}{\sqrt{2}}(\hat{i}-\hat{j})$

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(B) Let $d = d_1 \hat{i} + d_2 \hat{j} + d_3 \hat{k}$.Since $d$ is a unit vector, $d_1^2 + d_2^2 + d_3^2 = 1$ $(i)$.Given $a \cdot d = 0$, where $a = \hat{i} + \hat{j}$, we have $(\hat{i} + \hat{j}) \cdot (d_1 \hat{i} + d_2 \hat{j} + d_3 \hat{k}) = 0$, which implies $d_1 + d_2 = 0$, so $d_2 = -d_1$ (ii).Given $b \cdot (c 	imes d) = 0$, this is the scalar triple product $[b, c, d] = 0$.Calculating the determinant:$[b, c, d] = \begin{vmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ d_1 & d_2 & d_3 \end{vmatrix} = 0$.Expanding along the first row:$0(0 - d_2) - 1(d_3 - d_1) + 1(d_2 - 0) = 0$.$-d_3 + d_1 + d_2 = 0$.Substituting $d_2 = -d_1$ into this equation, we get $-d_3 + d_1 - d_1 = 0$, which implies $d_3 = 0$ (iii).Substituting (ii) and (iii) into $(i)$:$d_1^2 + (-d_1)^2 + 0^2 = 1 \Rightarrow 2d_1^2 = 1 \Rightarrow d_1 = \pm \frac{1}{\sqrt{2}}$.Thus, $d_1 = \pm \frac{1}{\sqrt{2}}$ and $d_2 = \mp \frac{1}{\sqrt{2}}$.Therefore, $d = \pm \frac{1}{\sqrt{2}}(\hat{i} - \hat{j})$.

Let $a=\hat{i}+\hat{j}$, $b=\hat{j}+\hat{k}$ and $c=\hat{i}+\hat{k}$. If $d$ is a unit vector such that $a \cdot d=0$ and $b \cdot(c \times d)=0$, then $d=$

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