Find the value of - for which the vectors a-3-2k- and b-3k- are perpendicular to each other-

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Standard IX

Mathematics

Applications of Cross Product

Find the valu...

Question

Find the value of $λ$ for which the vectors $\to a = 3^i +^j - 2^k$ and $\to b =^i + λ^j - 3^k$ are perpendicular to each other.

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Solution

As we know that if two vectrs are perpendicular then their dot product is $0$ .
Therefore,
$\to a \cdot \to b = 0$
$(3^i +^j - 2^k) \cdot (^i + λ^j - 3^k) = 0$
$3 + λ + 6 = 0$
$\Rightarrow λ = - 9$

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Find the value of $λ$ which will make the vectors $\to a, \to b and \to c$ coplanar, where $\to a = 2^i -^j +^k, \to b =^i + 2^j - 3^k and \to c = 3^i - λ^j + 5^k$ .

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and

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\to a =^i + 2^j - 3^k

and

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Q. If

\to a = 2^i + 2^j + 3^k, \to b = -^i + 2^j +^k

and

\to c = 3^i +^j

are such that

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Q. Find a unit vector perpendicular to each of the vectors

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where

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\to b =^i + 2^j - 2^k

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Find the value of λ for which the vectors →a=3^i+^j−2^k and →b=^i+λ^j−3^k are perpendicular to each other.

As we know that if two vectrs are perpendicular then their dot product is 0.Therefore,→a⋅→b=0(3^i+^j−2^k)⋅(^i+λ^j−3^k)=03+λ+6=0⇒λ=−9

Find the value of $λ$ for which the vectors $\to a = 3^i +^j - 2^k$ and $\to b =^i + λ^j - 3^k$ are perpendicular to each other.

As we know that if two vectrs are perpendicular then their dot product is $0$ .
Therefore,
$\to a \cdot \to b = 0$
$(3^i +^j - 2^k) \cdot (^i + λ^j - 3^k) = 0$
$3 + λ + 6 = 0$
$\Rightarrow λ = - 9$