# If a vector 2i+3j+8k is perpendicular to the vector 4j−4i+alpha k then the value of alpha is? solve without using the property of dot product.

If a vector $2i+3j+8k$ is perpendicular to the vector $4j-4i+\alpha k$ then the value of α is? solve without using the property of dot product.
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periasemdy
Since the two vectors are perpendicular, a less natural way would be think of Pythagoras theorem
Let $\stackrel{\to }{AB}=2i+3j+8k$ and $\stackrel{\to }{AC}=-4i+4j+\alpha k$
Since $\stackrel{\to }{AB}\perp \stackrel{\to }{AC}$, Therefore we have $\stackrel{\to }{BC}=\stackrel{\to }{AC}-\stackrel{\to }{AB}=-6i+j+\left(\alpha -8\right)k$
Also $B{C}^{2}=A{C}^{2}+A{B}^{2}$
$\left(-6{\right)}^{2}+\left(1{\right)}^{2}+\left(\alpha -8{\right)}^{2}=\left(16+16+{\alpha }^{2}\right)+\left(4+9+64\right)$
$\alpha =-0.5$
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4enevi
You can use the Pythagorean theorem. If you add two vectors, you get$6i-j+\left(8+\alpha \right)k$. Then we have
$36+1+\left(8+\alpha {\right)}^{2}=4+9+64+16+16+{\alpha }^{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}16\alpha =8$