If a→, b→ and c→ are three vectors such that 3a→+4b→+6c→=0→, a→=3, b→=3 and c→=4, then the value of -864(a→.b→+b→⋅c→+c→.a→)/6 is equal to

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MARKS Content Team · Accepted Answer

Correct answer: 1863. 3a→+4b→=-6c→⇒3a→+4b→^2=36c→^2 ⇒9a→^2+16b→^2+24a→.b→=36c→^2 ⇒81+144+24a→.b→=576 ⇒a→.b→=351/24 Similarly, 3a→+6c→=-4b→⇒3a→+6c→^2=16b→^2 ⇒9a→^2+36c→^2+36a→.c→=16b→^2 ⇒81+576+36a→.c→=144 ⇒a→.c→=-513/36 And, 6c→+4b→=-3a→⇒36c→|^2+16b→|^2+48b→⋅c→=9a→^2 ⇒576+144+48b→.c→=81⇒b→.c→=-213/16 Hence, (-864(a→.b→+b→.c→+c→.a→))/6=(-12636+12312+11502)/6=1863

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors such that $3 \vec{a} + 4 \vec{b} + 6 \vec{c} = \vec{0},$ $|\vec{a}| = 3, |\vec{b}| = 3$ and $|\vec{c}| = 4,$ then the value of $- 864 (\frac{\vec{a} . \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} . \vec{a}}{6})$ is equal to

Solution

More Vector Algebra Questions

If a→, b→ and c→ are three vectors such that 3a→+4b→+6c→=0→, a→=3, b→=3 and c→=4, then the value of -864a→.b→+b→⋅c→+c→.a→6 is equal to

Solution

More Vector Algebra Questions

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors such that $3 \vec{a} + 4 \vec{b} + 6 \vec{c} = \vec{0},$ $|\vec{a}| = 3, |\vec{b}| = 3$ and $|\vec{c}| = 4,$ then the value of $- 864 (\frac{\vec{a} . \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} . \vec{a}}{6})$ is equal to