If c-2 a-3 b- where -a-2 and -b-3 and a-b-6- also c- is coplanar with p -2 k- and q -2 -k- andperpendicular to r -2 k- then c can beA- 5-2-B- 5-2-C- 5-2-D- 5-2-

The correct option is B 5√(2)(î ĵ)|c|=√(4|a|^2+9|b|^2 12a·b) |c|=5 A c is coplanar with nbsp;p andq So c=lp+mq c=l(î+ĵ 2k̂)+m(î+ 2ĵ+k̂) c=(l+m)î+(l 2m)ĵ+( ...

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Standard XII

Mathematics

Inequalities Involving Modulus Function

If c⃗=2 a⃗-3 ...

Question

If $\to c = 2 \to a - 3 \to b$ where $| \to a | = 2$ and $| \to b | = 3$ and $\to a \cdot \to b = 6$ , also $\to c$ is coplanar with $\to p =^i +^j - 2^k$ and $\to q =^i - 2^j +^k$ and perpendicular to $\to r =^i +^j + 2^k$ then $\to c$ can be

\frac{5}{\sqrt{2}} (^i +^j)

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\frac{5}{\sqrt{2}} (^i -^j)

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\frac{5}{2} (^i -^j)

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\frac{5}{2} (^i +^j)

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Solution

The correct option is B $\frac{5}{\sqrt{2}} (^i -^j)$
$| \to c | = \sqrt{4 | \to a |^{2} + 9 | \to b |^{2} - 12 \to a \cdot \to b}$
$| \to c | = 5$

A $\to c$ is coplanar with $\to p$ and $\to q$

So $\to c = l \to p + m \to q$
$\to c = l (^i +^j - 2^k) + m (^i + - 2^j +^k)$
$\to c = (l + m)^i + (l - 2 m)^j + (- 2 l + m)^k)$
now this is perperndicular to $\to r =^i +^j + 2^k$
$\to c \cdot \to r = 0$
Taking dot product we get
$2 l = m$
$\to c = 3 l (^i -^j)$
$3 \sqrt{2} | l | = 5$
$l = \pm \frac{5}{3 \sqrt{2}}$
$\to c = \pm \frac{5}{\sqrt{2}} (^i -^j)$

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If →c=2→a−3→b where |→a|=2 and |→b|=3 and →a⋅→b=6, also →c is coplanar with →p=^i+^j−2^k and →q=^i−2^j+^k and perpendicular to →r=^i+^j+2^k then →c can be

If $\to c = 2 \to a - 3 \to b$ where $| \to a | = 2$ and $| \to b | = 3$ and $\to a \cdot \to b = 6$ , also $\to c$ is coplanar with $\to p =^i +^j - 2^k$ and $\to q =^i - 2^j +^k$ and perpendicular to $\to r =^i +^j + 2^k$ then $\to c$ can be