A vector of magnitude - 2 coplanar with the vectors - a - - -2k- and - b - -2- -k- and perpendicular to the vector - c - - -k- is1 - -k- 2 - -k- 3 - - 4 - -k-

The correct option is B 2 Let O⃗=A î+B +c k̂ #10; #10; |O⃗|=√(2)=√(A^2+B^2+C^2) #10; #10; amp;[c⃗·a⃗b⃗]=0 #10; (coplanar) amp; ...

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Standard XI

Mathematics

Standard Form of an Ellipse

A vector of m...

Question

A vector of magnitude $\sqrt{2}$ coplanar with the vectors $\to a =^i +^j + 2^k$ and $\to b =^i + 2^j +^k$ and perpendicular to the vector $\to c =^i +^j +^k$ is
(1) $-^i -^k$ (2) $^j -^k$ (3) $^i -^j$ (4) $^i +^k$

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Solution

The correct option is B 2
$Let \to O = A^i + B^ȷ + c^k$

$| \to O | = \sqrt{2} = \sqrt{A^{2} + B^{2} + C^{2}}$

$\begin{matrix} [\to c \cdot \to a \to b] = 0 (coplanar) \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} A & B & c 1 & 1 & 2 1 & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow A (1 - 4) - B (1 - 2) + c (2 - 1) = 0 \Rightarrow - 3 A + B + C = 0 \Rightarrow 3 A - B - C = 0 -(i) \end{matrix}$

And. $\to O ⊥^{γ} \to c$

$\to O \cdot \to c = 0$

$\Rightarrow A + B + C = 0 - (i i)$

By across multiplication, we got

$A = 0, B = 4 k, c = - 4 k$

$\sqrt{2} = \sqrt{0 + 16 k^{2} + 16 k^{2}}$

$\Rightarrow \sqrt{2} - \sqrt{32 k^{2}}$

Squaring both sides

$\Rightarrow 2 = 32 k^{2}$

$\Rightarrow k^{2} = \frac{1}{16}$

$k = \pm \frac{1}{4}$

$\Rightarrow$ $\to 0 =^ȷ -^k$ or $-^j +^k$

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A vector of magnitude √2 coplanar with the vectors →a=^i+^j+2^k and →b=^i+2^j+^k and perpendicular to the vector →c=^i+^j+^k is(1) −^i−^k (2) ^j−^k (3) ^i−^j (4) ^i+^k

A vector of magnitude $\sqrt{2}$ coplanar with the vectors $\to a =^i +^j + 2^k$ and $\to b =^i + 2^j +^k$ and perpendicular to the vector $\to c =^i +^j +^k$ is
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