Let u,v,w be such that ∣u∣=1,∣v∣=2,∣w∣=3. If the projection of v along u is equal to that of w along u, and v,w are perpendicular to each other, then ∣u−v+w∣ equals:
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Official solution
Given v⋅u=w⋅u \ldots (1) and v⋅w=0 \ldots (2):
\[|\vec u-\vec v+\vec w|^2 = |\vec u|^2+|\vec v|^2+|\vec w|^2-2\vec u\cdot\vec v-2\vec v\cdot\vec w+2\vec u\cdot\vec w = 1+4+9+0 = 14\]
using (1) to cancel −2u⋅v+2u⋅w. So ∣u−v+w∣=14.