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If bar(a) and bar(b) any two non-collinear vectors lying in the same plane, then prove that any vector bar(r) coplanar with them can be uniquely expressed as bar(r)=t_(1)bar(a)+t_(2)bar(b), where t_(1)andt_(2) are scalars. |
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Answer» Solution :Take any points O in the plane of `bar(a),bar(b)andbar(r)`. Represents the vectors `bar(a),bar(b)andbar(r)` by `bar(OA),bar(OB)andbar(OR)`. Take the points P on `bar(a)` and Q `bar(b)` such that OPRQ is a PARALLELOGRAM. Now `bar(OP)andbar(OA)` are collinear vectors. `:.` there exists a non - zero scalar `t_(1)` such that `bar(OP)=t_(a)*bar(OA)=t_(1)*bar(a)`. Also `bar(OQ)andbar(OB)` are collinear vectors. `:.` there exixts a non-zero scalar `t_(2)` such that `bar(OQ)=t_(2)*bar(OB)=t_(2)*bar(b)`. Now, by parallelogram law of addition of vectors, `bar(OR)=bar(OP)+bar(OQ)"":.bar(r)=t_(1)bar(a)+t_(2)bar(b)` Thus `bar(r)` expressed as linear combination `t_(1)bar(a)+t_(2)bar(b)` Uniqueness: LET, if possible, `bar(r)=t_(1)^(')bar(a)+t_(2)^(')bar(b)`, where `t_(1)^('),t_(2)^(')` are non-zero scalars. Then `t_(1)bar(a)+t_(2)bar(b)=t_(1)^(')bar(a)+t_(2)^(')bar(b)` `:.(t_(1)-t_(1)^('))bar(a)=-(t_(2)-t_(2)^('))bar(b)`. . . . (1) WE want to show that `t_(1)=t_(1)^(')andt_(2)=t_(2)^(')`. Suppose `t_(1)!=t_(1)^('),i.e.,t_(1)-t_(1)^(')!=0andt_(2)!=t_(2)^(')!=0`. Then dividing both sides of (1) by `t_(1)-t_(1)^(')`, we GET, `bar(a)=-((t_(2)-t_(2)^('))/(t_(1)-t_(1)^(')))bar(b)` This shows that the VECTOR `bar(a)` is a non-zero scalar MULTIPLE of `bar(b)`. `:.bar(a)andbar(b)` are collinear vectors. This is a contradiction, since `bar(a),bar(b)` are given to be non-collinear. `:.t_(1)=t_(1)^(')` Similarly, we can show tath `t_(2)=t_(2)^(')` This shows that `bar(r)` is uniquwly expressed as a linear combination `t_(1)bar(a)+t_(2)bar(b)`. 
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