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A vector of length 3 perpendicular to each of the vectors $3 \mathbf{i}+\mathbf{j}-4 \mathbf{k}$ and $6 \mathbf{i}+5 \mathbf{j}-2 \mathbf{k}$ is
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The correct answer is:
$2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}$
Let the vector is $x \mathbf{i}+y \mathbf{j}+\mathbf{z} \mathbf{k}$. Now according to the conditions, $\sqrt{x^2+y^2+z^2}=3 \Rightarrow x^2+y^2+z^2=9$
$6 x+5 y-2 z=0$ and $3 x+y-4 z=0$
[ it is perpendicular to both vectors, hence by $a_1 b_1+a_2 b_2+a_3 b_3=0$ ]
On solving the equation (i), (ii) and (iii), we get $x=2, \quad y=-2$ and $z=1$.
Therefore, the required vector is $2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}$.
Trick : By inspection, the vector $2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}$ is of length 3 and also perpendicular to the given vectors.
$6 x+5 y-2 z=0$ and $3 x+y-4 z=0$
[ it is perpendicular to both vectors, hence by $a_1 b_1+a_2 b_2+a_3 b_3=0$ ]
On solving the equation (i), (ii) and (iii), we get $x=2, \quad y=-2$ and $z=1$.
Therefore, the required vector is $2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}$.
Trick : By inspection, the vector $2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}$ is of length 3 and also perpendicular to the given vectors.
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