(i) Dot product of a vector with $\hat{i} + \hat{j} - 3 \hat{k}, \hat{i} + 3 \hat{j} - 2 \hat{k} and 2 \hat{i} + \hat{j} + 4 \hat{k}$ are 0, 5 and 8 respectively. Find the vector.
(ii) Dot products of a vector with vectors $\hat{i} - \hat{j} + \hat{k}, 2 \hat{i} + \hat{j} - 3 \hat{k} and \hat{i} + \hat{j} + \hat{k}$ are respectively 4, 0 and 2. Find the vector.

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Solution

$(i) Let a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k} be the required vector. Given that (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (\overset{\land}{i} + \overset{\land}{j} - 3 \overset{\land}{k}) = 0 \Rightarrow a + b - 3 c = 0 . . . (1) (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (\overset{\land}{i} + 3 \overset{\land}{j} - 2 \overset{\land}{k}) = 5 \Rightarrow a + 3 b - 2 c = 5 . . . (2) (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (2 \overset{\land}{i} + \overset{\land}{j} + 4 \overset{\land}{k}) = 5 \Rightarrow 2 a + b + 4 c = 8 . . . (3) Solving (1), (2) and (3), we get a = 1, b = 2, c = 1 So, a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k} = \overset{\land}{i} + 2 \overset{\land}{j} + \overset{\land}{k}$

$(i i) Let a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k} be the required vector. Given that (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (\overset{\land}{i} - \overset{\land}{j} + \overset{\land}{k}) = 4 \Rightarrow a - b + c = 4 . . . (1) (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (2 \overset{\land}{i} + \overset{\land}{j} - 3 \overset{\land}{k}) = 0 \Rightarrow 2 a + b - 3 c = 0 . . . (2) (a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}) . (\overset{\land}{i} + \overset{\land}{j} + \overset{\land}{k}) = 2 \Rightarrow a + b + c = 2 . . . (3) Solving (1), (2) and (3), we get a = 2; b = - 1; c = 1 So, a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k} = 2 \overset{\land}{i} - \overset{\land}{j} + \overset{\land}{k}$

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