If p -k- and q -2 -k- find a vector of magnitude 5 -3 units perpendicular to the vector q and coplanar with vectors p and q-

Let r⃗ =a î+bĵ +c k̂. As r⃗⊥q⃗ so, r⃗.q⃗=0 i.e., a 2b +c =0 ....(i) Also r⃗ is coplanar with vectors p⃗ and q⃗ so, [p⃗q⃗r⃗]=0 ∴ 1 amp; 1 amp; 1 1 am ...

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Byju's Answer

Standard XII

Mathematics

Angle between a Plane and a Line

If p =î+ĵ+k̂ ...

Question

If $\to p =^i +^j +^k$ and $\to q =^i - 2^j +^k$ , find a vector of magnitude $5 \sqrt{3}$ units perpendicular to the vector $\to q$ and coplanar with vectors $\to p$ and $\to q$ .

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Solution

Let $\to r = a^i + b^j + c^k .$
As $\to r ⊥ \to q$ so, $\to r . \to q = 0$ i.e., a -2b +c =0 ....(i)
Also $\to r$ is coplanar with vectors $\to p$ and $\to q$ so, $[\to p \to q \to r] = 0$
$∴ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & - 2 & 1 a & b & c \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow 3 a - 3 c = 0$ i.e., a = c ...(ii)
By (i) and (ii), we get: b = c
$∴$ the direction ratios of $\to r$ are a,b,c i.e., c,c,c i.e., 1,1,1.
So, $\to r =^i +^j +^k$
Now, the required vector has magnitude of $5 \sqrt{3}$ so, required vector is $5 \sqrt{3} \times \to r$
Therefore, required vector $= 5 \sqrt{3} \times \frac{\to r}{| \to r |} = 5 \sqrt{3} \times \frac{^i +^j +^k}{\sqrt{3}} = 5^i + 5^j + 5^k$ .

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