Question
If $a=i+j-k, b=i-j+k$ and $c$ is a unit vector perpendicular to the vector $a$ and coplanar with $a$ and $b$, then a unit vector $d$ perpendicular to both $a$ and $c$ is(A) $\frac{1}{\sqrt{6}}(2 i-j+k)$(B) $\frac{1}{\sqrt{2}}(i+j)$(C) $\frac{1}{\sqrt{2}}(j+k)$(D) $\frac{1}{\sqrt{2}}(i+k)$
Step 1
That is, $c = \alpha a + \beta b$. Substituting the values of $a$ and $b$ into the equation, we get $c = \alpha(i+j-k) + \beta(i-j+k)$. Simplifying this, we get $c = (\alpha + \beta)i + (\alpha - \beta)j + (\beta - \alpha)k$. Show more…
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