if the vectors( i^+j^+k^) and 3i^form two sides of a triangle then area of the triangle is
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In general, area (ΔΔ) of triangle with two (consecutive) sides A⃗ A→ & B⃗ B→ is given by following formula
Δ=12|A⃗ ×B⃗ |Δ=12|A→×B→|
hence, the area (ΔΔ) of triangle with sides 6i−4j+4k6i−4j+4k & −4i+2j−3k−4i+2j−3k
Δ=12|(6i−4j+4k)×(−4i+2j−3k)|Δ=12|(6i−4j+4k)×(−4i+2j−3k)|
=12|4i+2j−4k|=12|4i+2j−4k|
=1242+22+(−4)2−−−−−−−−−−−−−√=1242+22+(−4)2
=12(6)=12(6)
=3 unit2
In general, area (ΔΔ) of triangle with two (consecutive) sides A⃗ A→ & B⃗ B→ is given by following formula
Δ=12|A⃗ ×B⃗ |Δ=12|A→×B→|
hence, the area (ΔΔ) of triangle with sides 6i−4j+4k6i−4j+4k & −4i+2j−3k−4i+2j−3k
Δ=12|(6i−4j+4k)×(−4i+2j−3k)|Δ=12|(6i−4j+4k)×(−4i+2j−3k)|
=12|4i+2j−4k|=12|4i+2j−4k|
=1242+22+(−4)2−−−−−−−−−−−−−√=1242+22+(−4)2
=12(6)=12(6)
=3 unit2
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